Formula & theorem reference

Main inequality. Write |z+w|^2 using |\zeta|^2 = \zeta\bar\zeta for any complex number \zeta: |z+w|^2 = (z+w)\overline{(z+w)} = (z+w)(\bar z+\bar w) = z\bar z + z\bar w + w\bar z + w\bar w = |z|^2 + \big(z\bar w + \overline{z\bar w}\big) + |w|^2, using w\bar z = \overline{z\bar w}. Since \zeta+\bar\zeta = 2\,\mathrm{Re}(\zeta) for any complex \zeta, the middle term is 2\,\mathrm{Re}(z\bar w), so |z+w|^2 = |z|^2 + 2\,\mathrm{Re}(z\bar w) + |w|^2. For any complex number \zeta, \mathrm{Re}(\zeta) \le |\zeta| (the real part of a number never exceeds its modulus), applied to \zeta = z\bar w: \mathrm{Re}(z\bar w) \le |z\bar w| = |z||\bar w| = |z||w|. Substituting, |z+w|^2 \le |z|^2 + 2|z||w| + |w|^2 = \big(|z|+|w|\big)^2. Both |z+w| and |z|+|w| are nonnegative real numbers, so taking square roots preserves the inequality: |z+w|\le|z|+|w|.

Equality analysis. Equality holds throughout exactly when \mathrm{Re}(z\bar w) = |z||w|, i.e. when \mathrm{Re}(z\bar w) = |z\bar w|. For any complex number \zeta = a+bi, \mathrm{Re}(\zeta) = |\zeta| forces b=0 and a\ge0 — a real part equal to the full modulus can only happen for a nonnegative real number, since otherwise |\zeta| = \sqrt{a^2+b^2} > a = \mathrm{Re}(\zeta) whenever b\neq0 or a<0. So equality holds iff z\bar w is a nonnegative real number. If w=0 this holds trivially (with z\bar w = 0), matching "one of z,w is 0." If w\neq0, z\bar w\ge0 (real) is equivalent to z = \dfrac{z\bar w}{|w|^2}\,w being a nonnegative real multiple of w (multiply both sides by w/|w|^2, using \bar w w = |w|^2) — i.e. z and w point along the same ray from the origin.

Reverse triangle inequality. Apply the main inequality to z-w and w in place of z and w: |z| = |(z-w)+w| \le |z-w|+|w| \implies |z|-|w| \le |z-w|. Symmetrically, applying it to w-z and z: |w| = |(w-z)+z| \le |w-z|+|z| \implies |w|-|z| \le |w-z| = |z-w|, using |w-z|=|{-1}||z-w|=|z-w|. Combining the two, -|z-w| \le |z|-|w| \le |z-w|, which is exactly \big||z|-|w|\big| \le |z-w|.

Substitute x = i\theta into the Maclaurin series for e^x and use i^2=-1, i^3=-i, i^4=1 (the pattern repeats every four powers): e^{i\theta} = \sum_{n=0}^{\infty}\frac{(i\theta)^n}{n!} = 1 + i\theta - \frac{\theta^2}{2!} - i\frac{\theta^3}{3!} + \frac{\theta^4}{4!} + i\frac{\theta^5}{5!} - \cdots Collecting the terms with no factor of i and the terms with one factor of i into two separate sums gives e^{i\theta} = \underbrace{\left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots\right)}_{\cos\theta} + i\underbrace{\left(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots\right)}_{\sin\theta}. The first parenthesis is exactly the Maclaurin series for \cos\theta and the second is exactly the Maclaurin series for \sin\theta, so e^{i\theta} = \cos\theta + i\sin\theta.

This is mathematical induction, a proof technique that gets a full treatment of its own in Chapter 14 — worth previewing here in full, rather than waiting, since it's the only way to certify a pattern for every positive integer n at once rather than checking cases one at a time. Applied to the statement P(n): "[r\operatorname{cis}\theta]^n = r^n\operatorname{cis}(n\theta)."

Base case (n=1): [r\operatorname{cis}\theta]^1 = r\operatorname{cis}\theta = r^1\operatorname{cis}(1\cdot\theta). True. +++ Inductive hypothesis: assume P(k) holds for some integer k \ge 1 — that is, assume [r\operatorname{cis}\theta]^k = r^k\operatorname{cis}(k\theta). +++ Inductive step: show P(k+1) follows. Split off one more factor of z = r\operatorname{cis}\theta: [r\operatorname{cis}\theta]^{k+1} = [r\operatorname{cis}\theta]^k \cdot [r\operatorname{cis}\theta] +++ Apply the inductive hypothesis to the first factor, then multiply the two polar numbers using the product rule just proved — moduli multiply, arguments add: = \big[r^k\operatorname{cis}(k\theta)\big]\cdot\big[r\operatorname{cis}\theta\big] = r^k \cdot r\;\operatorname{cis}\big(k\theta+\theta\big) = r^{k+1}\operatorname{cis}\big((k+1)\theta\big) This is exactly P(k+1). +++ Base case and inductive step both hold, so by the Principle of Mathematical Induction, [r\operatorname{cis}\theta]^n = r^n\operatorname{cis}(n\theta) for every integer n \ge 1. \blacksquare

Every known proof leans on machinery outside this course — complex analysis (Liouville's theorem: a bounded entire function is constant, applied to 1/f), topology (a degree-n map wraps the plane around itself n times, so it must hit 0), or Galois theory. None of those tools are available yet, so the result is taken on faith here, the way it is in every first course that uses it. What this course can do with it: combined with the Factor Theorem, it guarantees a degree-n polynomial factors completely into n linear factors over the complex numbers — that consequence, the Linear Factorization Theorem, is used freely and does have its own proof in the registry.

Sketch for a region R that is both x-simple and y-simple (the general case patches finitely many such regions together, with the shared interior edges canceling since each is traversed once in each direction). Write R = \{(x,y) : a \le x \le b,\ g_1(x) \le y \le g_2(x)\}. Then \iint_R -\frac{\partial P}{\partial y}\, dA = -\int_a^b \big[P(x,g_2(x)) - P(x,g_1(x))\big]\, dx, by the single-variable Fundamental Theorem applied to the inner integral. The right side is exactly \oint_C P\,dx, splitting C into its bottom piece y=g_1(x) (traversed left to right) and top piece y=g_2(x) (traversed right to left, contributing the minus sign). The same argument with R described as y-simple gives \iint_R \frac{\partial Q}{\partial x}\, dA = \oint_C Q\,dy. Adding the two established identities gives the theorem.

Introduce a third point Q = (x_2, y_1) — same height as P_1, same sideways position as P_2. The segment P_1Q is horizontal with length |x_2 - x_1|, and the segment QP_2 is vertical with length |y_2 - y_1|. A horizontal segment meeting a vertical one always forms a right angle, so triangle P_1QP_2 is a right triangle with legs |x_2 - x_1| and |y_2 - y_1|, and its hypotenuse is exactly the segment P_1P_2 whose length d we want. The Pythagorean theorem gives d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2, and taking the (non-negative) square root of both sides gives the formula.

(\Rightarrow) Suppose \lim_{z\to z_0} f(z) = L. Given \varepsilon>0, choose \delta>0 so that 0<|z-z_0|<\delta forces |f(z)-L|<\varepsilon. For any (x,y) with 0<|(x,y)-(x_0,y_0)|<\delta (the same distance, whether measured in the plane or as a complex modulus), the real part of f(z)-L satisfies |u(x,y)-a| \le |f(z)-L| < \varepsilon, since the real part of a complex number never exceeds the number's own modulus. The same bound applies to |v(x,y)-b|. So both real limits hold, with the identical \delta that worked for f.

(\Leftarrow) Suppose u(x,y)\to a and v(x,y)\to b as (x,y)\to (x_0,y_0). Given \varepsilon>0, choose \delta_1>0 so that |u(x,y)-a|<\varepsilon/2 whenever 0<|(x,y)-(x_0,y_0)|<\delta_1, and \delta_2>0 so that |v(x,y)-b|<\varepsilon/2 whenever 0<|(x,y)-(x_0,y_0)|<\delta_2. Let \delta=\min(\delta_1,\delta_2). Then for 0<|z-z_0|<\delta, the triangle inequality gives |f(z)-L| = |(u-a)+i(v-b)| \le |u-a|+|v-b| < \tfrac{\varepsilon}{2}+\tfrac{\varepsilon}{2} = \varepsilon.

Both directions used nothing about f beyond its own definition as u+iv, so the equivalence holds for any complex function with real and imaginary parts u, v.

The point at infinity has no modulus of its own to plug into the ordinary limit definition, so "z\to\infty" is given a meaning by relating it to an ordinary limit at an ordinary point. The substitution w=1/z sends |z|\to\infty to w\to 0 (and z\to 0 to w\to\infty), so define \lim_{z\to\infty} f(z) = L to mean exactly \lim_{w\to 0} f(1/w) = L in the ordinary sense already established: for every \varepsilon>0 there is a \delta>0 such that 0<|w|<\delta forces |f(1/w)-L|<\varepsilon. Undoing the substitution (z=1/w, so |w|<\delta means |z|>1/\delta), this says precisely that f(z) stays within \varepsilon of L once |z| is large enough — the natural meaning of "f(z)\to L as z heads to infinity in any direction."

The second statement defines what it means for f itself to blow up: g(z) = 1/f(z) is close to 0 exactly when f(z) is large in modulus, so declaring \lim_{z\to z_0} f(z)=\infty to mean \lim_{z\to z_0} 1/f(z) = 0 reduces an "infinite limit" to an ordinary limit of a reciprocal function, the same trick used for real-variable infinite limits.

Injectivity on a small neighborhood (proof sketch). Since f'(z_0)\neq0 and f' is continuous near z_0 (an analytic function's derivative is again analytic, hence continuous), shrink to a disk U_0 around z_0 on which f' is nonzero throughout — possible since f'(z_0)\ne 0 and f' is not identically 0 near z_0, so its zeros (if any nearby) are isolated. This alone does not yet force injectivity on all of U_0 (unlike the real one-variable case, a nonvanishing complex derivative does not immediately rule out two distinct points sharing a value without a further argument), so shrink further to a disk U\subseteq U_0 centered at z_0 small enough that U contains no second solution of f(z)=f(z_0) (possible since those solutions are isolated too, f being nonconstant near z_0). The full argument that such a U can be chosen injective throughout — not just avoiding a second solution of f(z)=f(z_0), but of f(z)=w for every w near f(z_0) simultaneously — is the part only sketched here: it runs Rouché's Theorem exactly as in the Open Mapping Theorem's proof (open-mapping-theorem.md) to show f(z)-w has, for z ranging over U and w near f(z_0), the same number of zeros (with multiplicity) as f(z)-f(z_0) has in U, namely one (a simple zero, since f'(z_0)\ne0); a simple zero of f(z)-w means exactly one solution z\in U, not a repeated root standing in for several, which is what upgrades "some preimage exists" (the Open Mapping Theorem's conclusion) to "exactly one preimage exists" (injectivity). Granting that upgrade, f is injective on U.

V=f(U) is open and f^{-1} is continuous. Openness of V is the Open Mapping Theorem applied directly to f on U (nonconstant, since f'(z_0)\neq0 rules out f being locally constant). Continuity of f^{-1}:V\to U follows from injectivity plus openness: for any open U'\subseteq U, f(U') is open in V by the same open mapping argument restricted to U', and f(U') = (f^{-1})^{-1}(U'), so f^{-1} pulls open sets back to open sets, i.e. is continuous.

f^{-1} is complex-differentiable, with the stated formula. Fix w_0\in V and let z_0'=f^{-1}(w_0)\in U (writing z_0' to allow this argument to apply at any point of U, not just the original z_0). For w\in V, w\neq w_0, write z=f^{-1}(w), so z\neq z_0' (by injectivity) and w=f(z), w_0=f(z_0'). Then \frac{f^{-1}(w)-f^{-1}(w_0)}{w-w_0} = \frac{z-z_0'}{f(z)-f(z_0')} = \frac{1}{\dfrac{f(z)-f(z_0')}{z-z_0'}}. As w\to w_0, continuity of f^{-1} (previous paragraph) gives z=f^{-1}(w)\to z_0', and since f is analytic at z_0' with f'(z_0')\neq0, the difference quotient \dfrac{f(z)-f(z_0')}{z-z_0'} \to f'(z_0')\neq0. So the reciprocal above converges to 1/f'(z_0'), i.e. (f^{-1})'(w_0) = \lim_{w\to w_0}\frac{f^{-1}(w)-f^{-1}(w_0)}{w-w_0} = \frac{1}{f'(z_0')} = \frac{1}{f'\big(f^{-1}(w_0)\big)} exists. Since w_0\in V was arbitrary, f^{-1} is complex-differentiable, hence analytic, on V, with the stated derivative formula everywhere.

(a). The derivative f'(z_0) = \lim_{h\to 0} \dfrac{f(z_0+h)-f(z_0)}{h} is assumed to exist, meaning the limit is the same no matter how the complex increment h approaches 0. Take h real first, h = \Delta x. Then f'(z_0) = \lim_{\Delta x\to 0} \frac{u(x_0+\Delta x,y_0) - u(x_0,y_0)}{\Delta x} + i\lim_{\Delta x\to 0}\frac{v(x_0+\Delta x,y_0)-v(x_0,y_0)}{\Delta x} = u_x(x_0,y_0) + iv_x(x_0,y_0), which already shows u_x and v_x exist at (x_0,y_0).

Now take h purely imaginary, h = i\Delta y. Then f'(z_0) = \lim_{\Delta y\to 0} \frac{u(x_0,y_0+\Delta y) - u(x_0,y_0)}{i\Delta y} + i\lim_{\Delta y\to 0}\frac{v(x_0,y_0+\Delta y)-v(x_0,y_0)}{i\Delta y} = \frac{1}{i}u_y(x_0,y_0) + v_y(x_0,y_0) = v_y(x_0,y_0) - iu_y(x_0,y_0), using 1/i = -i. This shows u_y and v_y exist too.

Since f'(z_0) is a single number, the two expressions for it must agree: u_x + iv_x = v_y - iu_y. Matching real and imaginary parts gives u_x = v_y and v_x = -u_y, which is the same pair of equations written the other way around. The first computation above already exhibits f'(z_0) = u_x + iv_x.

(b). This direction needs more than the equations alone — a function can have partial derivatives satisfying them at a single point without being complex-differentiable there. The continuity hypothesis is what closes the gap: continuous first partials make u and v each real (Fréchet) differentiable at every point of D, meaning each can be approximated near a point by its linear (tangent-plane) part with an error that shrinks faster than the step size. Writing out that linear approximation for u and v together and substituting the Cauchy–Riemann equations turns the real 2-variable linearization into a single complex linear map of the increment h — precisely the statement that \lim_{h\to 0} \frac{f(z_0+h)-f(z_0)}{h} exists, with value u_x + iv_x. Since z_0 \in D was arbitrary, f is complex-differentiable at every point of D, i.e. analytic on D.

Write f = u+iv. Since f' \equiv 0 on D, the Cauchy–Riemann equations give f' = u_x + iv_x = 0 at every point, so u_x \equiv 0 and v_x \equiv 0 on D; and u_x = v_y, u_y = -v_x then force u_y \equiv 0 and v_y \equiv 0 as well. All four first partials of u and v vanish identically on D.

Fix a point z_0 \in D and let S be any horizontal or vertical segment inside D starting at z_0. Along that segment only one of x,y varies, and the corresponding partial derivative of u (and of v) is zero throughout the segment, so u and v — one-variable functions of arc length along S with zero derivative — are constant on S, equal to their value at z_0.

Because D is open and connected, any two points z_0, z_1 \in D can be joined by a polygonal path lying entirely in D made up of finitely many horizontal and vertical segments (this is the standard characterization of connectedness for open sets in the plane). Applying the previous paragraph to each segment in turn, u and v keep the same value from one endpoint of the path to the next, hence u(z_1) = u(z_0) and v(z_1) = v(z_0). Since z_0,z_1 were arbitrary, u and v are constant on all of D, so f = u+iv is constant on D.

For the forward direction, analyticity on D makes u and v differentiable as many times as needed, with continuous higher partials, and gives the Cauchy–Riemann equations u_x = v_y, u_y = -v_x throughout D. Differentiate the first equation with respect to x and the second with respect to y: u_{xx} = v_{yx}, \qquad u_{yy} = -v_{xy}. Because v's mixed second partials are continuous, they're equal: v_{yx} = v_{xy}. Adding the two displayed equations therefore gives u_{xx} + u_{yy} = v_{yx} - v_{xy} = 0. The same computation with the roles of u and v exchanged (differentiate u_x = v_y with respect to y and u_y=-v_x with respect to x, then subtract, using equality of u's mixed partials) gives v_{xx}+v_{yy}=0. So both u and v are harmonic on D.

For the converse, fix a basepoint z_0 = (x_0,y_0) \in D and, for any z=(x,y) \in D, define v(z) = \int_{z_0}^{z} \bigl(-u_y\,dx + u_x\,dy\bigr), the integral taken along any path in D from z_0 to z. This is the Cauchy–Riemann equations turned into a recipe: if f=u+iv is to be analytic, v must satisfy v_x = -u_y and v_y = u_x, and the displayed integral is exactly the line integral of the 1-form \omega = -u_y\,dx + u_x\,dy built from the right-hand sides of those two equations.

Before v can be called well defined, the integral must be shown independent of the path chosen — otherwise v(z) isn't a single number. By Green's Theorem, the integral of \omega around any closed curve \gamma bounding a region R \subset D equals \oint_\gamma \omega = \iint_R \Bigl(\partial_x(u_x) - \partial_y(-u_y)\Bigr)\,dA = \iint_R (u_{xx}+u_{yy})\,dA = 0, since u is harmonic. So \omega integrates to 0 around every closed curve that bounds a region contained in D — the statement that \omega is a closed form. This is where simple connectedness enters: D being simply connected means every closed curve in D bounds a region lying entirely inside D (it can be shrunk to a point without leaving D), so the vanishing just shown applies to every closed curve in D, not merely small ones. A closed 1-form with that property is exact: given two paths \gamma_1,\gamma_2 from z_0 to z, the concatenation of \gamma_1 with \gamma_2 reversed is a closed curve in D, so \int_{\gamma_1}\omega - \int_{\gamma_2}\omega = \oint_{\gamma_1 - \gamma_2}\omega = 0, meaning \int_{\gamma_1}\omega = \int_{\gamma_2}\omega. The integral defining v(z) therefore does not depend on the path chosen, and v is a well-defined function on D.

It remains to check v actually satisfies the Cauchy–Riemann equations paired with u. Fix z=(x,y)\in D and a small disk around it contained in D (possible since D is open). For h small, compute v(x+h,y)-v(x,y) by choosing, for the path to (x+h,y), the path to (x,y) followed by the horizontal segment from (x,y) to (x+h,y) — legal since the path choice doesn't affect the value just shown. On that horizontal segment dy=0, so v(x+h,y)-v(x,y) = \int_x^{x+h} -u_y(t,y)\,dt, and dividing by h and letting h\to 0 gives, by the Fundamental Theorem of Calculus (using that u_y is continuous, since u harmonic is automatically smooth), v_x(x,y) = -u_y(x,y). The symmetric computation, reaching (x,y+h) via the vertical segment from (x,y) where dx=0, gives v(x,y+h)-v(x,y) = \int_y^{y+h} u_x(x,s)\,ds, so v_y(x,y) = u_x(x,y). These are exactly the Cauchy–Riemann equations v_x=-u_y, v_y=u_x (equivalently u_x=v_y, u_y=-v_x) for the pair (u,v). Since u and v have continuous first partials satisfying the Cauchy–Riemann equations throughout D, f=u+iv is analytic on D, and v is the harmonic conjugate the converse claims.

Suppose toward a contradiction that f is analytic and nonconstant on D, yet some z_0 \in D has a neighborhood N \subseteq D with |f(z_0)| \ge |f(z)| for every z \in N; write M = |f(z_0)|.

The mean value property forces constant modulus nearby. Fix r>0 small enough that the closed disk \{|z-z_0|\le r\} lies in N. For each \rho \in (0,r], the mean value property — f(z_0) equals the average of f over the circle of radius \rho about z_0, itself a short consequence of the Cauchy Integral Formula — f(z_0) = \frac{1}{2\pi}\int_0^{2\pi} f(z_0+\rho e^{i\theta})\,d\theta gives M = |f(z_0)| = \left|\frac{1}{2\pi}\int_0^{2\pi} f(z_0+\rho e^{i\theta})\,d\theta\right| \le \frac{1}{2\pi}\int_0^{2\pi}|f(z_0+\rho e^{i\theta})|\,d\theta \le \frac{1}{2\pi}\int_0^{2\pi} M\,d\theta = M, the last inequality because every point z_0+\rho e^{i\theta} lies in N. Equality throughout forces the continuous, nonnegative function \theta \mapsto M - |f(z_0+\rho e^{i\theta})| to average to zero, hence to vanish identically. As \rho ranges over (0,r], this makes |f| \equiv M on the entire disk \Delta = \{|z-z_0|\le r\}.

Constant modulus forces constant value. Write f=u+iv, so u^2+v^2 \equiv M^2 on \Delta. Differentiating in x and y, then substituting the Cauchy–Riemann equations u_y=-v_x, v_y=u_x, gives the linear system uu_x + vv_x = 0, \qquad vu_x - uv_x = 0 for (u_x,v_x), with coefficient determinant -(u^2+v^2) = -M^2. If M=0, then f\equiv0 on \Delta outright. If M\neq0, the determinant is nonzero, forcing u_x=v_x=0 — and then u_y=v_y=0 too, by Cauchy–Riemann again — so f'\equiv0 throughout \Delta. Either way f is constant on \Delta, equal to f(z_0), by the Zero Derivative Implies Constant theorem (\Delta is connected).

Open-and-closed connectedness finishes the job. Let S = \{z\in D : f(z) = f(z_0)\}. S is closed in D, being the preimage of a single point under the continuous map f, and S\neq\varnothing since \Delta\subseteq S. S is also open in D: near any point of S, the same mean-value computation just run at z_0 — with that point standing in for z_0, using that f already equals the constant f(z_0) throughout a whole disk touching it — forces f to equal f(z_0) throughout a further whole disk around it (the identical mechanism the Identity Theorem's own proof uses to spread a zero outward, applied here to the zero set of f-f(z_0) instead of to f itself). This is exactly the open-and-closed connectedness argument behind the harmonic maximum principle: an equality pinned down on one neighborhood spreads outward, one overlapping disk at a time, to the entire connected domain. Since D is connected and S is a nonempty subset of D that is both open and closed in D, S=D. Hence f \equiv f(z_0) on all of D — contradicting the standing assumption that f is nonconstant, and completing the proof.

This proof runs the Maximum Principle (maximum-principle-harmonic.md) against -u rather than re-deriving the mean-value argument from scratch; a direct argument via analytic functions (composing with -1 times a harmonic conjugate) would also work but is heavier machinery than this needs.

-u is harmonic whenever u is. Harmonicity is the condition u_{xx}+u_{yy}=0 (Laplace's equation), and this condition is linear in u: if u_{xx}+u_{yy}=0 then (-u)_{xx}+(-u)_{yy} = -(u_{xx}+u_{yy}) = 0 too. -u is also continuous on \overline D and nonconstant on D exactly when u is.

Apply the maximum principle to -u. By the already-registered Maximum Principle, -u attains its maximum on \partial D, never at an interior point unless -u (equivalently u) is constant. Let M be that maximum of -u, attained at a boundary point \mathbf b.

Translate back to u. For every point \mathbf x in \overline D, -u(\mathbf x) \le M = -u(\mathbf b), so u(\mathbf x) \ge u(\mathbf b). That is, u(\mathbf b) is the smallest value u takes anywhere on \overline D — the minimum of u, attained at the boundary point \mathbf b. Since the maximum of -u was shown to occur only at boundary points (once u is nonconstant), the minimum of u occurs only at boundary points too.

Apply the theorem to h = f-g, which is analytic on D and vanishes on the given subset, so the consequence follows directly from the main statement; it remains to justify that one.

Every zero of an analytic function is either isolated (some punctured neighborhood of it contains no other zero) or a limit point of other zeros. The Taylor expansion of f centered at the limit point z_0 — valid since f is analytic there — decides which case occurs: if the expansion is not identically zero, its lowest-order nonzero term of some degree n dominates near z_0, forcing f(z) \neq 0 on a punctured neighborhood of z_0, which would make z_0 an isolated zero and contradicts the hypothesis that z_0 is a limit point of zeros. So the Taylor expansion at z_0 must be identically zero, meaning f \equiv 0 on some disk around z_0.

From there, the set of points in D where f and all its derivatives vanish is shown to be both open (by the same Taylor-expansion argument applied at each of its points) and closed in D (as an intersection of zero sets of continuous functions); since D is connected and this set contains z_0, it must be all of D. This "isolated zeros" machinery is developed rigorously in the power series chapter, where the Taylor expansion argument sketched above is made precise.

Define g on D^- by g(z) = \overline{f(\bar z)}. For z \in D^-, \bar z \in D^+, so this makes sense, and g is analytic on D^-: if w = \bar z and h\to 0, then \frac{g(z+h)-g(z)}{h} = \frac{\overline{f(\bar z+\bar h)} - \overline{f(\bar z)}}{h} = \overline{\left(\frac{f(w+\bar h)-f(w)}{\bar h}\right)} \xrightarrow{h\to 0} \overline{f'(w)}, so g'(z) exists and equals \overline{f'(\bar z)} — conjugating a convergent limit and conjugating its input are both continuous operations that commute with the limit, and \bar h \to 0 exactly when h \to 0.

On I itself, \bar z = z, and f is real-valued there, so g(z) = \overline{f(\bar z)} = \overline{f(z)} = f(z): the two pieces agree on the shared segment I, so they glue into one continuous function F on D, equal to f on D^+ \cup I and to g on D^- \cup I.

F is analytic separately on the open sets D^+ and D^-, and continuous across I; gluing two analytic pieces along a curve where the combined function is continuous is enough to conclude analyticity on the union, by a Morera-type argument (any small closed triangle straddling I has vanishing integral, built from the vanishing integrals on its two analytic halves plus a limiting argument for the strip touching I). That gluing step is stated informally here; it's made fully rigorous in Chapter 7 once Morera's Theorem and integral estimates are available. Granting it, F is analytic on all of D and extends f, and for z \in D^-, F(z) = g(z) = \overline{f(\bar z)}, which is the reflection identity F(\bar z) = \overline{F(z)} written out (check both directly on D^+, on D^-, and on I, where it reduces to the real-valued hypothesis).

Substitute x = i\theta into the Maclaurin series for e^x and use i^2=-1, i^3=-i, i^4=1 (the pattern repeats every four powers): e^{i\theta} = \sum_{n=0}^{\infty}\frac{(i\theta)^n}{n!} = 1 + i\theta - \frac{\theta^2}{2!} - i\frac{\theta^3}{3!} + \frac{\theta^4}{4!} + i\frac{\theta^5}{5!} - \cdots Collecting the terms with no factor of i and the terms with one factor of i into two separate sums gives e^{i\theta} = \underbrace{\left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots\right)}_{\cos\theta} + i\underbrace{\left(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots\right)}_{\sin\theta}. The first parenthesis is exactly the Maclaurin series for \cos\theta and the second is exactly the Maclaurin series for \sin\theta, so e^{i\theta} = \cos\theta + i\sin\theta.

(a). Write z_1=x_1+iy_1, z_2=x_2+iy_2. Then \exp(z_1)\exp(z_2) = e^{x_1}e^{x_2}(\cos y_1+i\sin y_1)(\cos y_2+i\sin y_2). Multiplying out the last factor and using i^2=-1 gives real part \cos y_1\cos y_2 - \sin y_1\sin y_2 = \cos(y_1+y_2) and imaginary part \sin y_1\cos y_2+\cos y_1\sin y_2 = \sin(y_1+y_2) by the real angle-addition formulas. Since e^{x_1}e^{x_2}=e^{x_1+x_2}, the product equals e^{x_1+x_2}(\cos(y_1+y_2)+i\sin(y_1+y_2)) = \exp(z_1+z_2).

(b). \exp(z+2\pi i) = e^x(\cos(y+2\pi)+i\sin(y+2\pi)) = e^x(\cos y+i\sin y) = \exp(z), since real \cos and \sin have period 2\pi. For minimality, suppose \exp(z_1)=\exp(z_2). By (c) below \exp(z_2)\neq 0, so dividing gives \exp(z_1-z_2) = \exp(z_1)/\exp(z_2) = 1 (division is multiplication by a reciprocal, and \exp(-z_2)=1/\exp(z_2) follows from (a) with z_1+z_2 replaced by 0: \exp(z_2)\exp(-z_2)=\exp(0)=1). Writing z_1-z_2=a+bi, \exp(a+bi)=e^a(\cos b+i\sin b)=1 forces e^a=1 (equal moduli, since |\cos b+i\sin b|=1), so a=0, and then \cos b=1,\sin b=0 forces b=2\pi k for some integer k.

(c). |\exp(z)| = |e^x(\cos y+i\sin y)| = e^x\cdot 1 = e^x > 0 for every real x, so \exp(z) can never equal 0.

(d). Write \exp(z)=u(x,y)+iv(x,y) with u=e^x\cos y, v=e^x\sin y. Then u_x=e^x\cos y=v_y and u_y=-e^x\sin y=-v_x, and all four partials are continuous everywhere — the Cauchy–Riemann equations hold on all of \mathbb{C} with continuous partials, so \exp is entire, and its derivative is \exp'(z)=u_x+iv_x=e^x\cos y+ie^x\sin y=\exp(z).

Write w=u+iv and solve \exp(w)=z, i.e. e^{u}(\cos v+i\sin v)=z. Taking the modulus of both sides, e^{u}\cdot 1 = |z| (since |\cos v+i\sin v|=1), and |z|>0 since z\neq 0, so u=\ln|z| — the ordinary real logarithm, uniquely determined.

Dividing both sides of e^u(\cos v+i\sin v)=z by e^u=|z| leaves \cos v+i\sin v = z/|z|, a number of modulus 1. This says exactly that v is a valid argument of z: every argument of z works (since adding any 2\pi k to a valid v doesn't change \cos v or \sin v), and by the exponential's periodicity (established above), any two solutions v_1,v_2 of \cos v+i\sin v = z/|z| differ by a multiple of 2\pi. So the full solution set for v is exactly \arg z, and every w=u+iv=\ln|z|+i\arg z constructed this way does satisfy \exp(w)=z — substituting back in and running the addition law confirms it, so no extra or missing solutions remain.

Write L_k := \ln|z| + i(\operatorname{Arg} z+2\pi k) for the k-th branch of \log z, so the k-th branch of z^c is \exp(cL_k). Two branches \exp(cL_{k_1}) and \exp(cL_{k_2}) are equal exactly when c(L_{k_1}-L_{k_2}) = c\cdot 2\pi i(k_1-k_2) is an integer multiple of 2\pi i (the exponential's periodicity), i.e. exactly when c(k_1-k_2)\in\mathbb{Z}.

(a) c=n: n(k_1-k_2) is automatically an integer for any integers k_1,k_2, so every branch coincides — call the common value z^n. It equals \exp(n\ln|z|+in\operatorname{Arg} z) = |z|^n(\cos(n\operatorname{Arg} z)+i\sin(n\operatorname{Arg} z)), which is exactly what repeated multiplication of z by itself n times produces (by the same angle-addition computation used in the exponential's own addition law).

(b) c=p/q, \gcd(p,q)=1: two branches k_1,k_2 coincide exactly when \frac{p}{q}(k_1-k_2)\in\mathbb{Z}, i.e. when q divides p(k_1-k_2); since \gcd(p,q)=1 this happens exactly when q divides k_1-k_2. So branches k and k+q always coincide, and no two branches strictly between them do — exactly q distinct values, one for each residue of k modulo q.

(c) If c\notin\mathbb{Q} (or c is not real), suppose two branches k_1\neq k_2 coincided: then c(k_1-k_2) would be a nonzero integer, forcing c to be the rational number \dfrac{c(k_1-k_2)}{k_1-k_2} — contradicting c irrational (and if c has nonzero imaginary part, c(k_1-k_2) has nonzero imaginary part too for any nonzero integer k_1-k_2, so it can never equal a real integer). So every branch is distinct, and there are infinitely many of them.

Linking identity. \sin(iz) = \dfrac{\exp(i\cdot iz)-\exp(-i\cdot iz)}{2i} = \dfrac{\exp(-z)-\exp(z)}{2i} = \dfrac{-1}{i}\cdot \dfrac{\exp(z)-\exp(-z)}{2} = i\sinh z, using -1/i=i. The other three linking identities follow by the identical substitution z\mapsto iz into the exponential definitions above.

\sin^2 z+\cos^2 z=1. Expand both squares directly from the definitions: \sin^2 z = \frac{\exp(2iz) - 2 + \exp(-2iz)}{-4}, \qquad \cos^2 z = \frac{\exp(2iz)+2+\exp(-2iz)}{4}. Adding the two (the first with the minus sign already absorbed since (2i)^2=-4): the \exp(2iz) and \exp(-2iz) terms cancel between the two fractions once the sign is tracked through, leaving \dfrac{2}{4}+\dfrac{2}{4}=1. This is the same algebraic identity that proves \cos^2\theta+\sin^2\theta=1 for real \theta — nothing about the computation used z being real, which is exactly why the identity survives unchanged.

\cosh^2 z - \sinh^2 z=1. Either repeat the same expansion with \exp(z),\exp(-z) in place of \exp(iz),\exp(-iz) (the minus sign between the two exponentials squared, \exp(-2z), no longer picks up the extra factor of -1 that i^2 contributed above, which is exactly why this identity has a minus sign where the trig one has a plus), or apply the two linking identities directly: \cosh^2 z-\sinh^2 z = \cos^2(iz) - (-i\sin(iz))^2 = \cos^2(iz)+\sin^2(iz) = 1.

\arcsin. Set w=\arcsin z, so \sin w=z: \frac{\exp(iw)-\exp(-iw)}{2i} = z. Let u=\exp(iw) (so \exp(-iw)=1/u). Clearing denominators, u - \dfrac1u = 2iz, i.e. u^2 - 2izu - 1 = 0, a quadratic in u. The quadratic formula gives u = \frac{2iz\pm\sqrt{(2iz)^2+4}}{2} = iz \pm \sqrt{1-z^2}. (The two signs are exactly the two values of the square root of 1-z^2 — a genuine two-valuedness, not an extra restriction, since either sign of u solves the same equation.) Taking \log of both sides of \exp(iw)=u gives iw = \log u, so w = \arcsin z = -i\log\big(iz+\sqrt{1-z^2}\big).

\arctan. Set w=\arctan z, so \tan w = z, i.e. \dfrac{\exp(iw)-\exp(-iw)}{i(\exp(iw)+\exp(-iw))}=z. With u=\exp(iw) again, \dfrac{u-1/u}{i(u+1/u)}=z \iff \dfrac{u^2-1}{i(u^2+1)}=z \iff u^2-1 = iz(u^2+1) \iff u^2(1-iz)=1+iz \iff u^2=\dfrac{1+iz}{1-iz}. Taking \log: 2iw=\log\!\left(\dfrac{1+iz}{1-iz}\right), so w=\arctan z = -\frac{i}{2}\log\!\left(\frac{1+iz}{1-iz}\right).

The other four (\arccos, \operatorname{arcsinh}, \operatorname{arccosh}, \operatorname{artanh}) come from the exact same substitution u=\exp(\pm w) turning the defining equation into a quadratic (or, for \operatorname{artanh}, a linear equation) in u — only the specific exponential definition being inverted changes.

Parametrize C by z(t), t\in[a,b], so that \int_C f(z)\,dz = \int_a^b f(z(t))z'(t)\,dt and L = \int_a^b |z'(t)|\,dt. For a complex-valued function g(t) of a real variable, the modulus of its integral never exceeds the integral of its modulus: \left|\int_a^b g(t)\,dt\right| \le \int_a^b |g(t)|\,dt, which follows by writing \int_a^b g(t)\,dt = re^{i\phi} for some r\ge0,\phi, so that r = e^{-i\phi}\int_a^b g(t)\,dt = \int_a^b e^{-i\phi}g(t)\,dt; since r is real, this equals its own real part, r = \int_a^b \operatorname{Re}(e^{-i\phi}g(t))\,dt \le \int_a^b |e^{-i\phi}g(t)|\,dt = \int_a^b |g(t)|\,dt, using \operatorname{Re}(w)\le|w| at every point of the integrand.

Applying this with g(t) = f(z(t))z'(t), \left|\int_C f(z)\,dz\right| = \left|\int_a^b f(z(t))z'(t)\,dt\right| \le \int_a^b |f(z(t))||z'(t)|\,dt \le \int_a^b M|z'(t)|\,dt = M\int_a^b|z'(t)|\,dt = ML, using |f(z(t))|\le M at every point of C in the middle step.

Antiderivative exists \implies path-independence. Suppose F'=f on D, and let C be any contour in D from z_1 to z_2, parametrized by z(t), t\in[a,b], with z(a)=z_1, z(b)=z_2. By the chain rule, \frac{d}{dt}F(z(t)) = F'(z(t))z'(t) = f(z(t))z'(t), so \int_C f(z)\,dz = \int_a^b f(z(t))z'(t)\,dt = \int_a^b \frac{d}{dt}F(z(t))\,dt = F(z(b)) - F(z(a)) = F(z_2)-F(z_1), the middle equality being the ordinary (real-variable) Fundamental Theorem of Calculus applied separately to the real and imaginary parts of F(z(t)). This value depends only on the endpoints z_1,z_2, not on which contour C joins them — path-independence. Taking z_1=z_2 (a closed contour) gives \oint_C f\,dz = F(z_1)-F(z_1)=0.

Path-independence \implies antiderivative exists. Fix a basepoint z_0\in D and, for each z\in D, define F(z) = \int_{z_0}^{z} f(w)\,dw, the integral taken over any contour in D from z_0 to z — this is well-defined (independent of which contour is chosen) precisely because of the path-independence hypothesis. It remains to check F'(z)=f(z) at an arbitrary z\in D. Since D is open, a small disk around z lies in D; for h small, path-independence lets F(z+h)-F(z) be computed by integrating f along the straight segment from z to z+h (rather than recomputing the whole path from z_0), so \frac{F(z+h)-F(z)}{h} - f(z) = \frac{1}{h}\int_{z}^{z+h} \big(f(w)-f(z)\big)\,dw, after writing f(z) = \frac{1}{h}\int_z^{z+h} f(z)\,dw (a constant integrated over a segment of length |h|). Continuity of f at z makes |f(w)-f(z)| small for w near z, and the ML inequality bounds the right side by that small quantity, times the segment's length |h|, divided by |h| — so the right side shrinks to 0 as h\to0. Hence F'(z)=f(z), and since z\in D was arbitrary, F is an antiderivative of f on D. (This differentiation-under-the-limit step is the same estimate used throughout this chapter; it is not elaborated further here.)

Main statement (proof sketch). The classical argument is due to Goursat and is genuinely more delicate than anything else in this course; only the idea is given here, not a fully rigorous version. Triangulate the region enclosed by C into a single triangle T (any simple closed contour can be built up from finitely many triangles, so it suffices to treat one). Bisecting each side of T subdivides it into four congruent sub-triangles, each with half the perimeter of T, whose boundary integrals of f sum exactly to \oint_T f(z)\,dz — the interior edges shared by adjacent sub-triangles are traversed once in each direction and cancel. By the pigeonhole principle, at least one sub-triangle T_1 has \left|\oint_{T_1} f\right| \ge \frac14\left|\oint_T f\right|. Repeating this quartering on T_1, then on the resulting sub-triangle, and so on, produces a nested sequence of shrinking triangles T\supset T_1\supset T_2\supset\cdots with perimeter of T_n equal to 2^{-n} times the perimeter of T, and \left|\oint_{T_n} f\right| \ge 4^{-n}\left|\oint_T f\right|. The nested triangles shrink to a single point z^* inside T (a compactness argument — a nested sequence of shrinking closed regions with diameters going to 0 has exactly one point common to all of them), where f is analytic. Analyticity at z^* means f(z) is well approximated near z^* by its linear part, f(z) = f(z^*) + f'(z^*)(z-z^*) + \varepsilon(z)(z-z^*) with \varepsilon(z)\to0 as z\to z^*; the constant and linear terms f(z^*)+f'(z^*)(z-z^*) have an antiderivative (a polynomial in z), so their contour integral around any closed triangle vanishes by the Fundamental Theorem for Contour Integrals, leaving only the error term. Bounding that error term with the ML inequality on T_n — whose perimeter and diameter are both shrinking geometrically like 2^{-n}, while \varepsilon(z)\to0 on T_n — shows \left|\oint_{T_n}f\right| is smaller than 4^{-n} times any fixed positive number, for n large enough. Comparing this to the lower bound above forces \oint_T f = 0. A fully rigorous version of this "shrink and estimate" argument needs more real-analysis machinery (uniform continuity on compact sets, in particular) than this course assumes; the geometric idea — that quartering shrinks the region's perimeter by half and its "defect" by at worst a quarter each time, so the defect cannot survive the limit — is the genuine content of the theorem.

Multiply connected corollary. Cut the region between C_0 and the inner contours C_1,\dots,C_n by finitely many connecting segments (crosscuts) so that the resulting boundary is a single simple closed contour on which f is analytic everywhere on and inside it; the main theorem gives 0 for the integral around this cut boundary. Each crosscut is traversed once in each direction, so those pieces cancel, leaving exactly \oint_{C_0} f\,dz (outer boundary, counterclockwise) minus \sum_k \oint_{C_k} f\,dz (inner boundaries, counterclockwise) — the minus sign because the cut construction traverses each inner hole clockwise as seen from inside the cut region. Setting this difference to 0 gives the stated identity.

Contour deformation corollary. Take n=1 in the multiply connected corollary, with C_0 the outer of the two contours and C_1 the inner one (the region between them is exactly the "annulus" swept out by deforming one contour into the other without crossing a point of non-analyticity), giving \oint_{C_0}f = \oint_{C_1}f directly.

(a). The integrand g(z) = f(z)/(z-z_0) is analytic on and inside C except at z_0 itself, so it is not covered directly by Cauchy–Goursat. Instead, let C_r be a small circle of radius r centered at z_0, small enough to lie entirely inside C; since g is analytic on the region between C and C_r, the contour-deformation corollary of Cauchy–Goursat gives \oint_C \frac{f(z)}{z-z_0}\,dz = \oint_{C_r} \frac{f(z)}{z-z_0}\,dz for every such r. Write f(z) = f(z_0) + \big(f(z)-f(z_0)\big) and split the right side into two integrals. For the first, \oint_{C_r} \frac{f(z_0)}{z-z_0}\,dz = f(z_0)\oint_{C_r}\frac{dz}{z-z_0} = f(z_0)\cdot 2\pi i, using the direct parametrization z=z_0+re^{i\theta}, dz = ire^{i\theta}\,d\theta, so \oint_{C_r}\frac{dz}{z-z_0} = \int_0^{2\pi} i\,d\theta = 2\pi i, independent of r. For the second, continuity of f at z_0 means that for any \varepsilon>0, |f(z)-f(z_0)|<\varepsilon once r is small enough that |z-z_0|=r stays inside that continuity neighborhood; the ML inequality then bounds \left|\oint_{C_r} \frac{f(z)-f(z_0)}{z-z_0}\,dz\right| \le \frac{\varepsilon}{r}\cdot(2\pi r) = 2\pi\varepsilon, which can be made arbitrarily small. Since the left side of the original deformation identity does not depend on r at all, while the right side equals 2\pi i f(z_0) plus a quantity that can be forced below any 2\pi\varepsilon, that quantity must in fact be 0. Therefore \oint_C f(z)/(z-z_0)\,dz = 2\pi i f(z_0), which rearranges to the stated formula.

(b). The same strategy proves the derivative formula, with one more layer of estimation: differentiating under the integral sign (justified by the same kind of difference-quotient and ML-inequality estimate used in part (a), applied to \frac{1}{z-z_0} and its powers rather than carried out in full here) shows that if f(z_0) is given by the integral in part (a), then f'(z_0) is given by differentiating the integrand in z_0 once, f''(z_0) by differentiating it twice, and so on, producing exactly the stated formula with (z-z_0)^{n+1} in the denominator and the constant n! appearing from the n-th derivative of (z-z_0)^{-1}. Because this differentiation can be repeated for every n, f has derivatives of every order at z_0.

Corollary. Part (b) shows every analytic f has an n-th derivative at every interior point of its domain, for every n; and each derivative f^{(n)}, being given itself by a Cauchy-type integral formula in z_0, is differentiable in z_0 by the same argument one order further — hence analytic.

Fix any point z_0\in\mathbb{C}. For every radius R>0, the derivative formula from the Cauchy Integral Formula (part (b), with n=1) applied to the circle C_R of radius R centered at z_0 gives f'(z_0) = \frac{1}{2\pi i}\oint_{C_R} \frac{f(z)}{(z-z_0)^2}\,dz. On C_R, |z-z_0|=R, so |f(z)/(z-z_0)^2| \le B/R^2 using the global bound |f(z)|\le B; the ML inequality (length of C_R is 2\pi R) then gives Cauchy's inequality for the first derivative, |f'(z_0)| \le \frac{1}{2\pi}\cdot\frac{B}{R^2}\cdot 2\pi R = \frac{B}{R}. This holds for every R>0, and f being entire means the formula above is valid no matter how large R is taken — there is no boundary to run into. Letting R\to\infty forces B/R\to0, so |f'(z_0)|=0, i.e. f'(z_0)=0. Since z_0\in\mathbb{C} was arbitrary, f'\equiv0 on all of \mathbb{C}, and \mathbb{C} is connected, so by the Zero Derivative Implies Constant theorem, f is constant.

Since f is nonvanishing on D, g = 1/f is defined at every point of D; it is analytic there too, being the quotient of the analytic function 1 by the analytic, nonvanishing function f (the quotient rule for analytic functions applies wherever the denominator doesn't vanish). g is also nonconstant on D: if g were constant, so would f=1/g be, contradicting the hypothesis.

The Maximum Modulus Principle (maximum-principle-analytic-functions.md) applies to g: since g is analytic and nonconstant on D, |g| has no interior maximum on D. But |g(z)| = \frac{1}{|f(z)|}, so a point z_0\in D where |f| has an interior minimum is exactly a point where |g| has an interior maximum (the reciprocal reverses the inequality: |f(z_0)|\le|f(z)| near z_0 if and only if |g(z_0)|\ge|g(z)| near z_0). Since |g| has no interior maximum, |f| has no interior minimum.

For the boundary consequence: |f| is continuous on the closed, bounded region R (as f is continuous there), so |f| attains a minimum somewhere on R by the Extreme Value Theorem. That minimum cannot occur at an interior point of R by the paragraph above, so it occurs on \partial R.

Fix z with |z-z_0|<R_0, and choose a radius r with |z-z_0|<r<R_0, so that the circle C_r=\{|w-z_0|=r\} lies inside the disk of analyticity and z lies strictly inside C_r. The Cauchy Integral Formula gives f(z) = \frac{1}{2\pi i}\oint_{C_r} \frac{f(w)}{w-z}\,dw. Rewrite the kernel using the geometric series. For w on C_r, \frac{1}{w-z} = \frac{1}{(w-z_0)-(z-z_0)} = \frac{1}{w-z_0}\cdot\frac{1}{1-\frac{z-z_0}{w-z_0}} = \frac{1}{w-z_0}\sum_{n=0}^{\infty}\left(\frac{z-z_0}{w-z_0}\right)^n, which is a valid geometric series expansion because \left|\frac{z-z_0}{w-z_0}\right| = \frac{|z-z_0|}{r} < 1 for every w on C_r — the ratio is a fixed number less than 1, independent of w, so the series converges uniformly in w on C_r. Substituting back, \frac{f(w)}{w-z} = \sum_{n=0}^{\infty} \frac{f(w)}{(w-z_0)^{n+1}}(z-z_0)^n, and the uniform convergence in w justifies integrating term by term around C_r: f(z) = \frac{1}{2\pi i}\oint_{C_r} \frac{f(w)}{w-z}\,dw = \sum_{n=0}^{\infty} \left(\frac{1}{2\pi i}\oint_{C_r} \frac{f(w)}{(w-z_0)^{n+1}}\,dw\right)(z-z_0)^n. Each coefficient in parentheses is exactly f^{(n)}(z_0)/n!, by the Cauchy Integral Formula's derivative-formula part (b) — the circle C_r may be used for every n since z_0 is inside it and f is analytic on and inside it. This gives f(z) = \sum_{n=0}^{\infty}\frac{f^{(n)}(z_0)}{n!}(z-z_0)^n, and since z was an arbitrary point of the disk \{|z-z_0|<R_0\} (with r chosen after z, only to be squeezed between |z-z_0| and R_0), the series converges to f(z) throughout the whole disk.

g is continuous. Uniform convergence on compact subsets of D means: for every z_0\in D, the f_n converge to g uniformly on some closed disk around z_0 contained in D. A uniform limit of continuous functions is continuous, so g is continuous on that disk; since z_0\in D was arbitrary, g is continuous on all of D.

Every closed contour integral of g vanishes. Let C be any closed contour lying in D; C is compact (it is the continuous image of a closed interval), so f_n\to g uniformly on C. Each f_n is analytic on D, so by the Cauchy–Goursat Theorem (cauchy-goursat-theorem.md), \oint_C f_n(z)\,dz = 0 \quad\text{for every } n. Uniform convergence on C justifies interchanging the limit in n with the contour integral: given \varepsilon>0, uniform convergence gives an N with |f_n(z)-g(z)|<\varepsilon for every z\in C once n\ge N, and the ML inequality then bounds \left|\oint_C f_n(z)\,dz - \oint_C g(z)\,dz\right| = \left|\oint_C \big(f_n(z)-g(z)\big)\,dz\right| \le \varepsilon\cdot\operatorname{length}(C), which shrinks to 0 as n\to\infty since \operatorname{length}(C) is a fixed number. So \oint_C f_n(z)\,dz \to \oint_C g(z)\,dz, and since the left side is identically 0 for every n, the limit forces \oint_C g(z)\,dz = 0.

Conclude via Morera's theorem. g is continuous on D (first step) and every closed contour integral of g over a contour in D vanishes (second step) — exactly the hypotheses of Morera's Theorem (morera-theorem.md), which concludes g is analytic on D.

Contrast with real variables. This theorem has no real-variable analog, and the contrast is worth stating plainly: a uniform limit of real-differentiable functions need not be differentiable at all. The standard example is a sequence of smooth functions approximating |x| arbitrarily closely and uniformly, whose limit fails to be differentiable at x=0 — uniform convergence preserves continuity in the real case but does nothing to control how wildly the functions oscillate at the scale their derivatives see. Complex differentiability is a far more rigid condition than real differentiability (an analytic function is automatically infinitely differentiable, equals its own Taylor series locally, and is entirely determined on a domain by its values on a small disk), and it is exactly this rigidity — packaged through Morera's theorem and the Cauchy–Goursat theorem behind it — that lets a merely uniform limit inherit full analyticity for free, with no separate control on the derivatives required at all.

Let f(z) denote the common value of the two series on \{|z-z_0|<r\}; this is an analytic function there (a convergent power series defines an analytic function on its disk of convergence), so it has derivatives of every order, and those derivatives may be computed by differentiating either series term by term any number of times within the disk.

Evaluating f itself at z=z_0 kills every term except the constant one in each series, giving a_0 = f(z_0) = b_0.

Differentiate both series once and evaluate at z=z_0: every term a_n(z-z_0)^n contributes na_n(z-z_0)^{n-1}, which vanishes at z=z_0 for every n\ge2 and equals a_1 for n=1, so f'(z_0) = a_1 from the left series and f'(z_0) = b_1 from the right, giving a_1=b_1.

In general, differentiating a power series k times term by term and evaluating at z=z_0 leaves only the n=k term (every lower-order term has been differentiated away to 0, and every higher-order term still carries a factor of (z-z_0) that vanishes at z=z_0), producing f^{(k)}(z_0) = k!\,a_k from one series and f^{(k)}(z_0) = k!\,b_k from the other. Since f^{(k)}(z_0) is a single number, k!\,a_k = k!\,b_k, and dividing by k!\neq0 gives a_k=b_k. As k\ge0 was arbitrary, a_n=b_n for every n.

This is a proof sketch rather than a fully worked-out argument. Fix z\in A and choose radii \rho_1,\rho_2 with r_1<\rho_1<|z-z_0|<\rho_2<r_2, so z lies in the sub-annulus between the circles C_{\rho_1} and C_{\rho_2}. Cutting this sub-annulus with a crosscut turns its boundary into a single simple closed contour on which f is analytic on and inside, so the Cauchy Integral Formula (applied to this cut contour, with the crosscut pieces canceling exactly as in the multiply connected corollary of Cauchy–Goursat) gives f(z) = \frac{1}{2\pi i}\oint_{C_{\rho_2}} \frac{f(w)}{w-z}\,dw - \frac{1}{2\pi i}\oint_{C_{\rho_1}} \frac{f(w)}{w-z}\,dw, both circles traversed counterclockwise, with a minus sign on the inner one exactly as in that corollary.

For the outer integral, |z-z_0|<\rho_2=|w-z_0| for w on C_{\rho_2}, so the same geometric-series expansion of \frac{1}{w-z} used in Taylor's Theorem applies, and term-by-term integration produces the non-negative-power part \sum_{n\ge0} c_n(z-z_0)^n with c_n as given above (using C_{\rho_2}, though contour deformation lets any C in the annulus be used instead).

For the inner integral, the roles reverse: |w-z_0|=\rho_1<|z-z_0| for w on C_{\rho_1}, so instead expand \frac{1}{w-z} = \frac{-1}{(z-z_0)-(w-z_0)} = \frac{-1}{z-z_0}\cdot\frac{1}{1-\frac{w-z_0}{z-z_0}} = -\sum_{n=0}^{\infty} \frac{(w-z_0)^n}{(z-z_0)^{n+1}}, valid since \left|\frac{w-z_0}{z-z_0}\right|=\rho_1/|z-z_0|<1 on C_{\rho_1} — the same geometric-series trick as before, with the roles of w-z_0 and z-z_0 swapped. Term-by-term integration of -\frac{1}{2\pi i}\oint_{C_{\rho_1}} f(w)\cdot\left(-\sum_n \frac{(w-z_0)^n}{(z-z_0)^{n+1}}\right)dw produces, after reindexing n\mapsto n-1, exactly the negative-power part \sum_{n\ge1} c_{-n}(z-z_0)^{-n} with c_{-n} matching the stated formula (again with C_{\rho_1} replaceable by any C in the annulus, by contour deformation, since the integrand is analytic throughout the annulus for every fixed n). Adding the two parts gives the full two-sided series. A fully rigorous version of the interchange between the sum and the integral in each piece uses the same uniform-convergence justification as in Taylor's Theorem, not repeated here.

By definition, z_0 is a pole of order m exactly when the Laurent series of f on a punctured disk around z_0 has c_{-m}\neq0 as its lowest-indexed nonzero coefficient, i.e. f(z) = \frac{c_{-m}}{(z-z_0)^m} + \frac{c_{-m+1}}{(z-z_0)^{m-1}} + \cdots + \frac{c_{-1}}{z-z_0} + c_0 + c_1(z-z_0)+\cdots with c_{-m}\neq0. Multiplying through by (z-z_0)^m gives \phi(z) := (z-z_0)^m f(z) = c_{-m} + c_{-m+1}(z-z_0) + \cdots + c_{-1}(z-z_0)^{m-1} + c_0(z-z_0)^m+\cdots, which is exactly a convergent power series in (z-z_0) on that same punctured disk, extendable to z_0 itself by the same series (a removable singularity), so \phi is analytic at z_0 with \phi(z_0)=c_{-m}\neq0. This proves the forward direction, and gives f(z)=\phi(z)/(z-z_0)^m. Conversely, if f(z)=\phi(z)/(z-z_0)^m with \phi analytic and nonzero at z_0, then \phi's own Taylor series \phi(z) = \sum_{k\ge0} \frac{\phi^{(k)}(z_0)}{k!}(z-z_0)^k divided by (z-z_0)^m produces a Laurent series for f whose lowest term is \phi(z_0)/(z-z_0)^m, nonzero since \phi(z_0)\neq0, so z_0 is a pole of order exactly m.

For the residue, \operatorname{Res}_{z=z_0} f = c_{-1}, the coefficient of (z-z_0)^{-1} in the Laurent series above — equivalently the coefficient of (z-z_0)^{m-1} in the power series for \phi shown just above. By the ordinary Taylor coefficient formula, that coefficient is \phi^{(m-1)}(z_0)/(m-1)!. When m=1 this reads \phi^{(0)}(z_0)/0! = \phi(z_0), matching the stated case split (the m>1 formula already covers m=1 in this form, but the simple pole case is recorded separately since it is by far the most common).

Apply the theorem to h = f-g, which is analytic on D and vanishes on the given subset, so the consequence follows directly from the main statement; it remains to justify that one.

Every zero of an analytic function is either isolated (some punctured neighborhood of it contains no other zero) or a limit point of other zeros. The Taylor expansion of f centered at the limit point z_0 — valid since f is analytic there — decides which case occurs: if the expansion is not identically zero, its lowest-order nonzero term of some degree n dominates near z_0, forcing f(z) \neq 0 on a punctured neighborhood of z_0, which would make z_0 an isolated zero and contradicts the hypothesis that z_0 is a limit point of zeros. So the Taylor expansion at z_0 must be identically zero, meaning f \equiv 0 on some disk around z_0.

From there, the set of points in D where f and all its derivatives vanish is shown to be both open (by the same Taylor-expansion argument applied at each of its points) and closed in D (as an intersection of zero sets of continuous functions); since D is connected and this set contains z_0, it must be all of D. This "isolated zeros" machinery is developed rigorously in the power series chapter, where the Taylor expansion argument sketched above is made precise.

Surround each singular point z_j with a small circle C_j, centered at z_j, with radius small enough that the C_j lie inside C, pairwise disjoint, and each containing no singular point of f other than z_j itself. Then f is analytic on the region bounded outside by C and inside by C_1,\dots,C_k — exactly the multiply connected setting of the Cauchy–Goursat theorem's corollary — so \oint_C f(z)\,dz = \sum_{j=1}^{k} \oint_{C_j} f(z)\,dz, each C_j traversed counterclockwise. It remains only to identify each \oint_{C_j} f\,dz.

By definition, the residue \operatorname{Res}_{z=z_j} f is the coefficient c_{-1} of (z-z_j)^{-1} in f's Laurent series on a punctured disk around z_j — and C_j, lying inside that punctured disk (having been chosen small enough), is exactly the kind of circle Laurent's Theorem uses to recover coefficients: c_{-1} = \frac{1}{2\pi i}\oint_{C_j} \frac{f(w)}{(w-z_j)^{-1+1}}\,dw = \frac{1}{2\pi i}\oint_{C_j} f(w)\,dw, taking n=-1 in Laurent's coefficient formula. Rearranged, this says \oint_{C_j} f\,dz = 2\pi i\operatorname{Res}_{z=z_j} f directly, for each j. Substituting into the sum above gives \oint_C f(z)\,dz = \sum_{j=1}^{k} 2\pi i\operatorname{Res}_{z=z_j} f = 2\pi i\sum_{j=1}^{k}\operatorname{Res}_{z=z_j} f.

This is a derivation of why the formula is built this way, not a proof in the usual sense — the right-hand side is a definition, and what needs justifying is the \frac{1}{w^2} factor and the minus sign, so that the resulting quantity deserves to be called a residue "at infinity" at all.

Pick R large enough that every finite singularity of f lies inside |z|<R, and parametrize the circle |z|=R by z=Re^{i\theta}, \theta:0\to2\pi (counterclockwise). Substituting w=1/z, so dz=-\frac{1}{w^2}\,dw, gives w=\frac1R e^{-i\theta} — as \theta runs from 0 to 2\pi, the argument of w runs from 0 down to -2\pi, so this same path traces the circle |w|=1/R once clockwise, not counterclockwise. (The map z\mapsto 1/z is conformal, but it swaps which side of the circle counts as "inside" — the unbounded region |z|>R becomes the bounded disk |w|<1/R — which is exactly why the induced loop runs the opposite way.)

So, purely as a change of variable in the same integral, \oint_{|z|=R,\,\text{ccw}} f(z)\,dz = \oint_{|w|=1/R,\,\text{cw}} f\!\left(\frac1w\right)\left(-\frac{1}{w^2}\right)dw = \oint_{|w|=1/R,\,\text{ccw}} \frac{1}{w^2}f\!\left(\frac1w\right)dw, the last step reversing the clockwise loop to counterclockwise, which flips the sign of the integral and cancels the minus sign already sitting on dz. The right-hand contour encloses only w=0 (once R is large enough that the transformed integrand has no other singularity inside |w|<1/R), so by Cauchy's Residue Theorem this equals 2\pi i\operatorname{Res}_{w=0}\left[\frac{1}{w^2}f(1/w)\right].

On the other hand, the same left-hand integral is computed directly by Cauchy's Residue Theorem on the z-side: since |z|=R encloses every finite singularity of f, \oint_{|z|=R,\,\text{ccw}} f(z)\,dz = 2\pi i\,S, where S is the sum of all of f's finite residues.

Equating the two expressions for the same integral gives S = \operatorname{Res}_{w=0}\left[\frac{1}{w^2}f(1/w)\right]. Defining \operatorname{Res}_{z=\infty}f to be the negative of that quantity is exactly what makes S+\operatorname{Res}_{z=\infty}f=0 hold — a finite residue and the residue at infinity are built so that every singularity of f on the whole extended plane, near or far, contributes to one clean total of zero.

Choose R large enough that |z_j|<R for every j. By Cauchy's Residue Theorem, \oint_{|z|=R} f(z)\,dz = 2\pi i\sum_{j=1}^{k}\operatorname{Res}_{z=z_j}f, the contour traversed counterclockwise. By the derivation behind the definition of the residue at infinity, this same integral also equals -2\pi i\operatorname{Res}_{z=\infty} f. Equating the two expressions for one integral and dividing by 2\pi i gives \sum_{j=1}^{k}\operatorname{Res}_{z=z_j}f = -\operatorname{Res}_{z=\infty}f, which is the claimed identity rearranged.

By definition, z_0 is a pole of order m exactly when the Laurent series of f on a punctured disk around z_0 has c_{-m}\neq0 as its lowest-indexed nonzero coefficient, i.e. f(z) = \frac{c_{-m}}{(z-z_0)^m} + \frac{c_{-m+1}}{(z-z_0)^{m-1}} + \cdots + \frac{c_{-1}}{z-z_0} + c_0 + c_1(z-z_0)+\cdots with c_{-m}\neq0. Multiplying through by (z-z_0)^m gives \phi(z) := (z-z_0)^m f(z) = c_{-m} + c_{-m+1}(z-z_0) + \cdots + c_{-1}(z-z_0)^{m-1} + c_0(z-z_0)^m+\cdots, which is exactly a convergent power series in (z-z_0) on that same punctured disk, extendable to z_0 itself by the same series (a removable singularity), so \phi is analytic at z_0 with \phi(z_0)=c_{-m}\neq0. This proves the forward direction, and gives f(z)=\phi(z)/(z-z_0)^m. Conversely, if f(z)=\phi(z)/(z-z_0)^m with \phi analytic and nonzero at z_0, then \phi's own Taylor series \phi(z) = \sum_{k\ge0} \frac{\phi^{(k)}(z_0)}{k!}(z-z_0)^k divided by (z-z_0)^m produces a Laurent series for f whose lowest term is \phi(z_0)/(z-z_0)^m, nonzero since \phi(z_0)\neq0, so z_0 is a pole of order exactly m.

For the residue, \operatorname{Res}_{z=z_0} f = c_{-1}, the coefficient of (z-z_0)^{-1} in the Laurent series above — equivalently the coefficient of (z-z_0)^{m-1} in the power series for \phi shown just above. By the ordinary Taylor coefficient formula, that coefficient is \phi^{(m-1)}(z_0)/(m-1)!. When m=1 this reads \phi^{(0)}(z_0)/0! = \phi(z_0), matching the stated case split (the m>1 formula already covers m=1 in this form, but the simple pole case is recorded separately since it is by far the most common).

Since q(z_0)=0 and q'(z_0)\neq0, z_0 is a simple (order-1) zero of q, so q factors as q(z)=(z-z_0)h(z) for some function h analytic at z_0 — divide q(z) by z-z_0 using q's own Taylor series at z_0, whose constant term is q(z_0)=0, so z-z_0 divides it exactly. Comparing derivatives at z_0: q'(z) = h(z) + (z-z_0)h'(z), so q'(z_0)=h(z_0).

Substituting, f(z) = \dfrac{p(z)}{(z-z_0)h(z)} = \dfrac{\phi(z)}{z-z_0} where \phi(z):=p(z)/h(z) is analytic at z_0 (since h(z_0)=q'(z_0)\neq0) and \phi(z_0)=p(z_0)/h(z_0)\neq0 (since p(z_0)\neq0). This is exactly the pole-order-and-residue formula's m=1 case, with residue \phi(z_0): \operatorname{Res}_{z=z_0} f = \phi(z_0) = \frac{p(z_0)}{h(z_0)} = \frac{p(z_0)}{q'(z_0)}.

Parametrize \Gamma_R by z=Re^{i\theta}, \theta\in[0,\pi], so dz = iRe^{i\theta}\,d\theta. Since iaz = iaR\cos\theta - aR\sin\theta, |e^{iaz}| = |e^{iaR\cos\theta}|\cdot|e^{-aR\sin\theta}| = e^{-aR\sin\theta}, using that e^{iaR\cos\theta} has modulus 1 (a purely imaginary exponent) and \sin\theta\ge0 on [0,\pi] makes -aR\sin\theta real. Hence \left|\int_{\Gamma_R} f(z)\,dz\right| \le \int_0^{\pi} |g(Re^{i\theta})|\,e^{-aR\sin\theta}\, R\,d\theta \le M(R)\,R\int_0^{\pi} e^{-aR\sin\theta}\,d\theta. The graph of \sin\theta on [0,\pi] is symmetric about \theta=\pi/2, so the integral splits evenly: \int_0^{\pi} e^{-aR\sin\theta}\,d\theta = 2\int_0^{\pi/2} e^{-aR\sin\theta}\,d\theta. Now the key estimate: on [0,\pi/2], the concave graph of \sin\theta lies on or above the straight chord joining (0,0) to (\pi/2,1), which has equation y=\frac{2}{\pi}\theta — that is, \sin\theta \ge \frac{2}{\pi}\theta \qquad \text{for } 0\le\theta\le\pi/2. (Concavity: \sin''\theta=-\sin\theta\le0 on [0,\pi/2], so the curve never dips below any chord connecting two of its own points, in particular this one.) Since e^{-x} is decreasing, this gives e^{-aR\sin\theta} \le e^{-2aR\theta/\pi}, so 2\int_0^{\pi/2} e^{-aR\sin\theta}\,d\theta \le 2\int_0^{\pi/2} e^{-2aR\theta/\pi}\,d\theta = 2\cdot\frac{\pi}{2aR}\Big(1-e^{-aR}\Big) < \frac{\pi}{aR}. Combining, \left|\int_{\Gamma_R} f(z)\,dz\right| < M(R)\,R\cdot\frac{\pi}{aR} = \frac{\pi M(R)}{a}. As R\to\infty, M(R)\to0 by hypothesis, so the right side tends to 0, and hence so does \int_{\Gamma_R} f(z)\,dz.

The two forms are equivalent. Along C, f is analytic and never zero, so f'/f is analytic on C, and (locally) f'/f is the derivative of a branch of \log f(z) = \ln|f(z)| + i\arg f(z). Summing these local antiderivatives around all of C, \oint_C (f'/f)\,dz equals the total change, once around C, of \ln|f(z)|+i\arg f(z). Since |f(z)| is single-valued and returns to its starting value when z returns to its starting point, the net change in \ln|f(z)| is 0; only \arg f(z), tracked continuously rather than reduced modulo 2\pi, can come back different from where it started (by a multiple of 2\pi), contributing i\,\Delta_C\arg f(z). So \oint_C(f'/f)\,dz = i\,\Delta_C\arg f(z), which rearranges to show the two displayed formulas say the same thing.

Proving the integral form via the residue theorem. f'/f is analytic inside and on C except at the zeros and poles of f, which are finitely many (a meromorphic function's poles are isolated, and its zeros are isolated as long as f is not identically 0, which the nonvanishing hypothesis on C rules out). By the residue theorem, \oint_C \frac{f'(z)}{f(z)}\,dz = 2\pi i\sum \operatorname{Res}\frac{f'}{f}, the sum over the zeros and poles of f inside C. It remains to compute each residue.

At a zero z_0 of order m, write f(z)=(z-z_0)^m\phi(z) with \phi analytic and \phi(z_0)\neq0 (the pole-order formula's defining form, here for a zero instead). Then f'(z) = m(z-z_0)^{m-1}\phi(z) + (z-z_0)^m\phi'(z), so \frac{f'(z)}{f(z)} = \frac{m(z-z_0)^{m-1}\phi(z)+(z-z_0)^m\phi'(z)}{(z-z_0)^m\phi(z)} = \frac{m}{z-z_0} + \frac{\phi'(z)}{\phi(z)}. The second term is analytic at z_0 (since \phi(z_0)\neq0), so f'/f has a simple pole at z_0 with residue exactly m, the order of the zero.

At a pole z_0 of order p, write f(z)=\psi(z)/(z-z_0)^p with \psi analytic and \psi(z_0)\neq0. Then f'(z) = \frac{\psi'(z)}{(z-z_0)^p} - \frac{p\,\psi(z)}{(z-z_0)^{p+1}}, so \frac{f'(z)}{f(z)} = \frac{\psi'(z)/(z-z_0)^p - p\psi(z)/(z-z_0)^{p+1}}{\psi(z)/(z-z_0)^p} = \frac{\psi'(z)}{\psi(z)} - \frac{p}{z-z_0}, again analytic in its first term at z_0, so f'/f has a simple pole at z_0 with residue -p, the negative of the order of the pole.

Summing these residues over every zero (contributing its order, positively) and every pole (contributing its order, negatively) inside C gives \sum\operatorname{Res}(f'/f) = Z - P. Substituting into the residue theorem's identity above, \oint_C \frac{f'(z)}{f(z)}\,dz = 2\pi i(Z-P), which rearranges to the stated integral formula.

The hypothesis |g(z)|<|f(z)| on C forces f(z)\neq0 on C (if f(z)=0 for some z\in C, the inequality would read |g(z)|<0, impossible), and it also forces (f+g)(z)\neq0 on C: if f(z)+g(z)=0 for some z\in C, then g(z)=-f(z), so |g(z)|=|f(z)|, contradicting the strict inequality. So both f and f+g are analytic and nonzero on C, and the argument principle applies to each (with no poles, since both are analytic — not just meromorphic — inside C), giving Z_f = \frac{1}{2\pi i}\oint_C \frac{f'(z)}{f(z)}\,dz, \qquad Z_{f+g} = \frac{1}{2\pi i}\oint_C \frac{f'(z)+g'(z)}{f(z)+g(z)}\,dz, where Z_f, Z_{f+g} denote the number of zeros inside C (with multiplicity) of f and f+g respectively.

For t\in[0,1], let f_t = f+tg, interpolating from f_0=f to f_1=f+g. On C, |tg(z)| \le |g(z)| < |f(z)| for every t\in[0,1], so by the same two-line argument as above, f_t(z)\neq0 on C for every t. Hence the argument principle applies to each f_t on C, and N(t) := \frac{1}{2\pi i}\oint_C \frac{f_t'(z)}{f_t(z)}\,dz is, for each fixed t, an integer — the zero count of f_t inside C. The integrand \frac{f_t'(z)}{f_t(z)} = \frac{f'(z)+tg'(z)}{f(z)+tg(z)} depends continuously on t for z\in C (the denominator never vanishes there, for any t\in[0,1], as just shown), and C has finite length, so N(t) is a continuous function of t\in[0,1] — a small change in t produces only a small change in the integral, by the ML inequality applied to the difference of integrands over the fixed contour C. A continuous, integer-valued function on the connected interval [0,1] must be constant (an integer-valued function taking two different values would have to pass through a non-integer value in between, by the intermediate value theorem, which is impossible). So N(0)=N(1), i.e. Z_f = N(0) = N(1) = Z_{f+g}.

It suffices to show: for every z_0\in D and w_0=f(z_0), some disk centered at w_0 lies entirely in f(D) — applied with D replaced by any open U\subseteq D (the same argument runs unchanged on U), this shows every image point of an open set is an interior point of the image, i.e. the image is open.

Isolate z_0 as the only nearby solution of f(z)=w_0. Since f is nonconstant and analytic on the connected domain D, the zeros of f(z)-w_0 are isolated (the Identity Theorem: a nonconstant analytic function's zero set has no accumulation point in D). So some r>0 exists with the closed disk \overline{\Delta} = \{|z-z_0|\le r\} contained in D and containing no zero of f(z)-w_0 other than z_0 itself. In particular f(z)-w_0\neq0 on the circle C=\{|z-z_0|=r\}, and since C is compact and |f(z)-w_0| is continuous and positive on it, it attains a positive minimum there: let m = \min_{z\in C} |f(z)-w_0| > 0.

Every target within m of w_0 is hit. Fix any w with |w-w_0|<m. On C, |(w_0-w)| < m \le |f(z)-w_0|, so writing F(z)=f(z)-w_0 and g(z)=w_0-w (a constant, hence analytic), |g(z)|<|F(z)| at every point of C. Rouché's Theorem (rouches-theorem.md) applies: F and F+g = f(z)-w have the same number of zeros, counted with multiplicity, inside C. F=f-w_0 has at least one zero inside C, namely z_0, so F+g=f-w has at least one zero inside C too — some z_1 inside C with f(z_1)=w. Hence w\in f(D).

Conclusion. Every w with |w-w_0|<m lies in f(D): the whole disk of radius m about w_0 is contained in f(D). Since z_0\in D (equivalently w_0\in f(D)) was arbitrary, every point of f(D) is an interior point, so f(D) is open. The identical argument with D replaced by any open U\subseteq D (the Identity Theorem step and the minimum-modulus step both only use analyticity on U and connectedness of the small disk \overline\Delta\subseteq U, not of all of D) shows f(U) is open too.

Why nonconstant is necessary. A constant function sends every open set to a single point, which is never open in \mathbb C (unless \mathbb C itself were a single point) — the theorem is specifically about the rigidity nonconstant analyticity forces.

Set up the Bromwich contour. For R>0 large enough that every pole of F lies inside it, let C_R be the closed contour built from the vertical segment L_R running from \gamma-iR up to \gamma+iR, closed by the left semicircular arc \Gamma_R=\{\gamma+Re^{i\theta}: \theta\in[\pi/2,3\pi/2]\} of radius R centered at \gamma. Traversed this way (L_R upward, then \Gamma_R through the left side back down to the start), C_R is positively oriented: every pole of F, having real part less than \gamma, eventually sits strictly inside C_R once R exceeds the distance from \gamma to the farthest pole.

Apply the residue theorem. For such R, \oint_{C_R} F(s)e^{st}\,ds = 2\pi i\sum_k \operatorname{Res}_{s=s_k}\big[F(s)e^{st}\big], a fixed number (for fixed t) independent of R once every pole is enclosed. Splitting the left side, \oint_{C_R} = \int_{L_R} + \int_{\Gamma_R}.

The arc vanishes -- by Jordan's lemma, rotated. Fix t>0 and substitute s=\gamma+iz (a rotation of the plane by 90° about \gamma, so ds=i\,dz). As s traces \Gamma_R with \theta\in[\pi/2,3\pi/2], writing \theta=\phi+\pi/2 shows z=-i(s-\gamma) traces exactly the UPPER semicircular arc \{z=Re^{i\phi}:\phi\in[0,\pi]\} (this is the same rotation identity Jordan's lemma's own upper-half-plane picture (jordans-lemma.md) is stated for, just relabeled). Define g(z)=F(\gamma+iz): since every pole s_k of F has \operatorname{Re}(s_k)<\gamma, the corresponding pole of g sits at z_k=-i(s_k-\gamma) with \operatorname{Im}(z_k)=-\operatorname{Re}(s_k-\gamma)>0 -- strictly in the open upper half-plane, hence outside the arc once R is large -- and g(z)=F(\gamma+iz)\to0 as |z|\to\infty because F(s)\to0 as |s|\to\infty by hypothesis. So g meets Jordan's lemma's own hypotheses exactly, with a=t>0: \int_{\Gamma_R} F(s)e^{st}\,ds = ie^{\gamma t}\int_{0}^{\pi} g(Re^{i\phi})\,e^{itRe^{i\phi}}\,iRe^{i\phi}\,d\phi \Big/ i = ie^{\gamma t}\int_{\Gamma_R'} e^{itz}g(z)\,dz \longrightarrow 0 as R\to\infty, where \Gamma_R' is the image upper arc -- this is literally Jordan's lemma's own conclusion, transported through the rotation s=\gamma+iz.

Take the limit. With \int_{\Gamma_R}\to0, the residue-theorem identity gives \int_{L_R} F(s)e^{st}\,ds = 2\pi i\sum_k \operatorname{Res}_{s=s_k}\big[F(s)e^{st}\big] - \int_{\Gamma_R} F(s)e^{st}\,ds \longrightarrow 2\pi i\sum_k \operatorname{Res}_{s=s_k}\big[F(s)e^{st}\big] as R\to\infty. But \int_{L_R}F(s)e^{st}\,ds = \int_{\gamma-iR}^{\gamma+iR}F(s)e^{st}\,ds is exactly the truncation of the Bromwich integral, so its limit as R\to\infty is \int_{\gamma-i\infty}^{\gamma+i\infty}F(s)e^{st}\,ds by definition. Dividing both sides by 2\pi i: \frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty} F(s)e^{st}\,ds = \sum_k \operatorname{Res}_{s=s_k}\big[F(s)e^{st}\big].

Bijectivity. The inverse map T^{-1}(z) = \frac{dz-b}{-cz+a} is again a Möbius transformation, since (-c)b - a(-d)\cdot(-1) — more directly, its own determinant is d\cdot a - (-b)(-c) = ad-bc \neq 0. Direct substitution confirms T^{-1}(T(z)) = z and T(T^{-1}(z)) = z for every z \in \hat{\mathbb{C}}, using the stated conventions at \infty and at the pole -d/c. A map with a two-sided inverse on \hat{\mathbb{C}} is a bijection of \hat{\mathbb{C}}.

Factorization. If c = 0, then ad \neq 0 and T(z) = \frac{a}{d}z + \frac{b}{d} is an affine map: a rotation-scaling by a/d followed by a translation by b/d. If c \neq 0, polynomial division gives T(z) = \frac{az+b}{cz+d} = \frac{a}{c} + \frac{bc-ad}{c(cz+d)}, so T factors as T = T_4 \circ T_3 \circ T_2 \circ T_1, \qquad T_1(z) = cz+d,\quad T_2(z) = \frac{1}{z},\quad T_3(z) = \frac{bc-ad}{c}\,z,\quad T_4(z) = z + \frac{a}{c}, a composition of a rotation-scaling-then-translation (T_1), the inversion T_2(z) = 1/z, another rotation-scaling (T_3), and a translation (T_4). Every Möbius transformation with c\neq 0 is therefore built from translations, rotation-scalings, and one inversion; the c=0 case uses only the first two kinds. So it suffices to prove property 2 for each of these three elementary kinds of map.

Translations and rotation-scalings preserve generalized circles. A translation z \mapsto z+\beta is a rigid motion of the plane: it sends an ordinary circle of center z_0, radius r to the ordinary circle of center z_0+\beta, radius r, and sends a line to a parallel line, hence a generalized circle to a generalized circle (both also fix \infty, matching the point that completes a line into a generalized circle). A rotation-scaling z \mapsto \alpha z (\alpha \neq 0) similarly sends the circle |z-z_0| = r to |w - \alpha z_0| = |\alpha| r (substitute z = w/\alpha), an ordinary circle, and sends a line through the origin's rotation and dilation to another line; again \infty \mapsto \infty.

Inversion preserves generalized circles. This is the one case that needs a genuine computation. Every generalized circle in the plane — ordinary circle or line — is the solution set of an equation of the form A(x^2+y^2) + Bx + Cy + E = 0, \qquad A,B,C,E \in \mathbb{R},\ (A,B,C)\neq(0,0,0), which is a circle when A \neq 0 (completing the square, provided B^2+C^2 > 4AE so the radius is real and positive) and a line when A = 0. Writing z = x+iy, note x^2+y^2 = z\bar z, x = \frac{z+\bar z}{2}, y = \frac{z-\bar z}{2i}, so the equation becomes A z\bar z + \frac{B}{2}(z+\bar z) + \frac{C}{2i}(z-\bar z) + E = 0, i.e. Az\bar z + \bar\beta z + \beta \bar z + E = 0 for the complex constant \beta = \frac{B}{2} - \frac{C}{2i} (equivalently \beta = \tfrac{B+iC}{2}). Now substitute z = 1/w (so \bar z = 1/\bar w) — this is precisely testing which points w have z = 1/w lying on the original generalized circle: A\cdot\frac{1}{w\bar w} + \bar\beta\cdot\frac1w + \beta\cdot\frac{1}{\bar w} + E = 0. Multiplying through by w\bar w (nonzero for w \neq 0) gives A + \bar\beta \bar w + \beta w + E w\bar w = 0, \quad\text{i.e.}\quad E w\bar w + \beta w + \bar\beta \bar w + A = 0, which has exactly the same form as the original equation, with the roles of A and E swapped and \beta \leftrightarrow \bar\beta (a relabeling that doesn't change the real coefficients B,C up to sign). So the image under z \mapsto 1/z of the generalized circle Az\bar z + \bar\beta z + \beta\bar z + E=0 is again a set of exactly this form — a generalized circle, with the coefficient pair (A,E) swapped to (E,A). The swap correctly tracks what happens to \infty and to 0 under z \mapsto 1/z: the original passes through z=0 exactly when E=0, and z=0 maps to w=\infty; the image equation Ew\bar w + \beta w + \bar\beta\bar w + A = 0 is a line (passes through \infty) exactly when its leading coefficient E is 0 — the same condition. So a generalized circle through the origin maps to a line, a generalized circle avoiding the origin maps to an ordinary circle, and this matches the swap of A and E in every case. In every case, inversion carries generalized circles to generalized circles.

Composition finishes property 2. Since T is a composition of maps each sending generalized circles to generalized circles, T itself sends generalized circles to generalized circles.

Three-point determination. Given distinct z_1,z_2,z_3, the map C(z) = \frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)} is Möbius (its numerator and denominator, expanded, give ad-bc \neq 0 because z_1,z_2,z_3 are distinct) and sends z_1\mapsto 0, z_2 \mapsto 1, z_3\mapsto\infty (interpreting the formula at z=z_3 via the \infty convention). Let C' be the analogous map built from w_1,w_2,w_3. Then T := C'^{-1}\circ C is Möbius (a composition of Möbius maps, using invertibility from part 1) and sends z_1,z_2,z_3 to w_1,w_2,w_3 respectively. For uniqueness, suppose T_1,T_2 both do this; then S = T_2^{-1}\circ T_1 is Möbius and fixes z_1,z_2,z_3. A Möbius map S(z) = \frac{az+b}{cz+d} fixing three distinct points satisfies cz^2+(d-a)z-b=0 for three distinct values of z; a nonzero quadratic has at most two roots, so c=0,\,d=a,\,b=0, forcing S(z)=z, the identity. Hence T_1=T_2.

Since f is analytic at z_0, it is differentiable there, so f(z_0+h) = f(z_0) + f'(z_0)\,h + \varepsilon(h)\,h, \qquad \varepsilon(h) \to 0 \text{ as } h \to 0, by the definition of the complex derivative — the error term is smaller than |h| itself in the limit, uniformly in the direction of h since h is a single complex number, not a pair of independent real parameters. This is the sense in which f is well approximated near z_0 by the affine map h \mapsto f(z_0) + f'(z_0)h: for h small, f(z_0+h) - f(z_0) \approx f'(z_0)h, with relative error going to 0 regardless of h's direction.

Multiplication by f'(z_0) is a rotation-scaling. Write f'(z_0) = \rho e^{i\alpha} in polar form, \rho = |f'(z_0)| > 0 (using the hypothesis f'(z_0)\neq 0). For any nonzero h = re^{i\phi}, f'(z_0)\,h = \rho e^{i\alpha}\cdot re^{i\phi} = \rho r\, e^{i(\phi+\alpha)}. So multiplying by f'(z_0) scales every vector's length by the fixed factor \rho and rotates every vector's direction by the fixed angle \alpha, regardless of the vector's own direction \phi — this is precisely what "acts like a rotation by \arg f'(z_0) combined with a scaling by |f'(z_0)|" means, and it holds uniformly in \phi, which is exactly why the local picture near z_0 looks like a single rigid-motion-plus-dilation rather than a direction-dependent distortion.

Angle preservation. Let \gamma_1,\gamma_2 be smooth curves through z_0, parametrized so \gamma_1(0)=\gamma_2(0)=z_0, with nonzero tangent vectors \gamma_1'(0) = h_1, \gamma_2'(0) = h_2. The angle between the curves at z_0 is, by definition, the angle between these tangent vectors, \theta = \arg(h_2/h_1) = \arg h_2 - \arg h_1. By the chain rule, the image curves have tangent vectors at f(z_0) equal to (f\circ\gamma_1)'(0) = f'(z_0)\,h_1, \qquad (f\circ\gamma_2)'(0) = f'(z_0)\,h_2, both nonzero since f'(z_0)\neq 0 and h_1,h_2\neq 0. The angle between the image tangent vectors is \arg\!\left(\frac{f'(z_0)h_2}{f'(z_0)h_1}\right) = \arg\!\left(\frac{h_2}{h_1}\right) = \theta, the factor f'(z_0) canceling exactly because it is the same nonzero complex number multiplying both tangent vectors — this is the algebraic heart of conformality: a single fixed nonzero scalar multiplying two vectors changes neither the angle between them nor which one is rotated into the other by turning counterclockwise. So the image curves meet at f(z_0) at the same angle \theta, measured in the same sense (from image of \gamma_1 to image of \gamma_2), as the original curves at z_0: f is conformal at z_0.

Both conclusions — the local rotation-scaling picture and angle preservation — trace back to the same fact: to first order, f near z_0 is translation by f(z_0) composed with multiplication by the single nonzero constant f'(z_0), and multiplication by a nonzero constant is exactly a rotation-plus-scaling, which by its uniformity in direction preserves every angle between directions through the origin.

Define g:\mathbb{D}\to\mathbb{C} by g(z) = \begin{cases} \dfrac{f(z)}{z}, & z\neq 0,\\[4pt] f'(0), & z=0.\end{cases} Since f is analytic on \mathbb{D} with f(0)=0, its Taylor series at 0 has no constant term, f(z) = a_1 z + a_2 z^2 + \cdots with a_1 = f'(0), so f(z)/z = a_1 + a_2 z + \cdots for z\neq 0 extends to an analytic function on all of \mathbb{D} agreeing with this power series — that extension is exactly g, and the singularity of f(z)/z at z=0 is removable.

Fix 0<r<1. On the circle |z|=r, |f(z)| \le 1 since f maps into \mathbb{D}, so |g(z)| = \frac{|f(z)|}{|z|} \le \frac{1}{r} \qquad (|z|=r). By the Maximum Modulus Principle, an analytic function on the closed disk |z|\le r attains its maximum modulus on the boundary circle, so |g(z)| \le \frac{1}{r} \qquad \text{for all } |z| \le r. This holds for every r<1; fixing any z\in\mathbb{D} and letting r\to 1^- (valid once r>|z|) gives |g(z)| \le 1 \qquad \text{for every } z\in\mathbb{D}. Unwinding the definition of g: for z\neq 0, |f(z)/z|\le 1, i.e. |f(z)|\le|z|, which is the first inequality (and holds trivially at z=0 too); and |g(0)| = |f'(0)| \le 1, the second inequality.

Equality case. Suppose |f(z_1)|=|z_1| for some z_1\neq 0, or |f'(0)|=1. In either case |g(z_1')| = 1 for the relevant point (z_1' = z_1 in the first case, z_1'=0 in the second) — an interior point of \mathbb{D} where |g| attains the value 1, which is also the supremum of |g| on \mathbb{D} just established. An analytic function whose modulus attains an interior maximum is constant, by the Maximum Modulus Principle's equality case: g(z) = c for all z \in \mathbb{D}, for some constant c with |c|=1 (since |g(z_1')|=1). Unwinding again, f(z) = z\,g(z) = cz for every z\in\mathbb{D}, as claimed.

The disk automorphisms. For fixed a \in \mathbb{D}, the map \varphi_a(z) = \dfrac{z-a}{1-\bar a z} is a Möbius transformation (mobius-transformations.md), since its coefficients \begin{pmatrix}1 & -a\\ -\bar a & 1\end{pmatrix} have determinant 1-|a|^2 \neq 0 for a\in\mathbb{D}. It sends a \mapsto 0, and one checks |\varphi_a(z)|<1 whenever |z|<1 directly: 1-|\varphi_a(z)|^2 = \frac{|1-\bar a z|^2 - |z-a|^2}{|1-\bar a z|^2} = \frac{(1-|a|^2)(1-|z|^2)}{|1-\bar a z|^2} > 0 for |z|<1,|a|<1 (expand both squared moduli in terms of z,\bar z,a,\bar a to see the middle equality). So \varphi_a maps \mathbb{D} into \mathbb{D}, and since \varphi_a is its own inverse up to sign/rotation — direct computation gives \varphi_a(\varphi_a(z)) = z when a is real, and in general \varphi_{-a} composed suitably inverts it — \varphi_a is in fact a bijection of \mathbb{D} onto itself, an automorphism. Composing \varphi_a with a rotation z\mapsto e^{i\theta}z (itself an automorphism of \mathbb{D}, since rotations preserve |z|<1) gives every map of the stated form e^{i\theta}\varphi_a(z), each one an automorphism of \mathbb{D}, being a composition of two automorphisms.

Reduction to Schwarz's Lemma. Fix z_1,z_2\in\mathbb{D} and let w_1 = f(z_1). Define g = \varphi_{w_1} \circ f \circ \varphi_{z_1}^{-1} = \varphi_{w_1}\circ f \circ \varphi_{-z_1}, using \varphi_{z_1}^{-1} = \varphi_{-z_1} (both are automorphisms fixing the relevant points inversely, by the computation above). Then g:\mathbb{D}\to\mathbb{D} is analytic, being a composition of analytic self-maps of \mathbb{D}, and g(0) = \varphi_{w_1}(f(\varphi_{-z_1}(0))) = \varphi_{w_1}(f(z_1)) = \varphi_{w_1}(w_1) = 0. So Schwarz's Lemma (schwarz-lemma.md) applies to g: |g(\zeta)|\le|\zeta| for every \zeta\in\mathbb{D}. Set \zeta = \varphi_{z_1}(z_2), so that \varphi_{-z_1}(\zeta) = z_2 and hence, unwinding the definition of g, g(\zeta) = \varphi_{w_1}\bigl(f(\varphi_{-z_1}(\zeta))\bigr) = \varphi_{w_1}(f(z_2)). The inequality |g(\zeta)|\le|\zeta| reads exactly \left|\varphi_{w_1}(f(z_2))\right| \le \left|\varphi_{z_1}(z_2)\right|, \quad\text{i.e.}\quad \left|\frac{f(z_2)-f(z_1)}{1-\overline{f(z_1)}f(z_2)}\right| \le \left|\frac{z_2-z_1}{1-\bar z_1 z_2}\right|, which is the claimed inequality (the two sides are unchanged in absolute value by swapping which point is "z_1" vs "z_2" up to conjugating the denominator, so the labeling is immaterial).

Equality case. By the equality case of Schwarz's Lemma, equality |g(\zeta)|=|\zeta| at the particular \zeta above forces g(z) = cz on all of \mathbb{D} for some |c|=1 — a rotation. Then f = \varphi_{w_1}^{-1}\circ g \circ \varphi_{z_1} = \varphi_{-w_1}\circ(c\cdot(\cdot))\circ\varphi_{z_1}, a composition of automorphisms of \mathbb{D} (translation-type \varphi's and a rotation), hence itself an automorphism of \mathbb{D}, and by the classification above must have the stated form e^{i\theta}\varphi_a(z) for some \theta,a. Conversely, if f already has that form, f is one of the automorphisms \varphi_a composed with a rotation, and running the argument backward — or noting directly that \varphi_a conjugates the quantity \left|\frac{f(z_1)-f(z_2)}{1-\overline{f(z_1)}f(z_2)}\right| into \left|\frac{z_1-z_2}{1-\bar z_1 z_2}\right| exactly, since both sides are built to be invariant under precomposing/postcomposing with disk automorphisms of this \varphi_a form — gives equality for every pair z_1,z_2, not merely some pair.

This card gives a plausibility explanation for why the exponents \alpha_k - 1 are the right choice, not a rigorous proof — showing that a specific f of this form actually maps onto a given polygon P (choosing the right x_k,A,B to match prescribed vertices w_k) requires the existence half of the Riemann Mapping Theorem together with boundary correspondence results beyond what a short derivation can supply; that is a genuine theorem, not just a formula check, and sits past this course's scope.

Why each factor supplies the right angle turn. Consider the integrand g(\zeta) = \prod_k (\zeta-x_k)^{\alpha_k-1}, so that f'(z) = A\,g(z). As z moves along the real axis through a point x that is not one of the x_k, each factor (z-x_k)^{\alpha_k-1} is a fixed nonzero complex number varying continuously, so \arg f'(z) = \arg A + \sum_k(\alpha_k-1)\arg(z-x_k) changes continuously — meaning the direction of the tangent vector to the image curve f(\mathbb{R}), which is f'(z) pointing along the direction of travel, turns continuously too. There is no corner in the image as long as z avoids the x_k: the boundary of P is traced out as a straight segment there, consistent with f mapping the segment between consecutive x_k's to the straight edge between the corresponding w_k's.

The interesting event happens as z crosses one particular x_k along the real axis. Near x_k, the single factor (z-x_k)^{\alpha_k-1} dominates the local behavior of \arg g(z) (the other factors (z-x_j)^{\alpha_j-1}, j\neq k, vary continuously and contribute no jump at z=x_k), and as z passes from just left of x_k to just right of x_k along the real axis, \arg(z-x_k) jumps by -\pi (the point z - x_k swings from the negative real axis, argument \pi, to the positive real axis, argument 0, a change of -\pi, or equivalently the point z-x_k passes through 0 with the ray from it sweeping clockwise by \pi as seen from the real line). This contributes a jump of (\alpha_k-1)\cdot(-\pi) = (1-\alpha_k)\pi to \arg g(z), hence to \arg f'(z), hence — since \arg f'(z) tracks the direction the boundary curve is heading — to the direction of travel along the image boundary. A tangent direction turning by (1-\alpha_k)\pi at a single point is exactly the exterior angle turn at a polygon vertex whose interior angle is \alpha_k\pi (interior angle plus exterior angle turn equals \pi for a straight continuation, so the turn is \pi - \alpha_k\pi = (1-\alpha_k)\pi). So the exponent \alpha_k - 1 is exactly calibrated to produce a corner of interior angle \alpha_k\pi in the image at the point w_k = f(x_k), and no exponent choice other than \alpha_k-1 (up to the additive freedom that doesn't affect the jump, since only the jump in argument at x_k matters) produces that specific turn.

Since f restricted to the real axis (away from the x_k) traces straight segments and turns by exactly the prescribed exterior angle at each x_k, the image of \mathbb{R} under f is, plausibly, the boundary of the polygon P with the prescribed vertex angles — this is the sense in which the formula is constructive: given the target angles, the exponents in the integrand are dictated immediately, while the finitely many remaining constants (A, B, and typically the positions of two of the x_k, which may be normalized/fixed by the group of Möbius self-maps of \mathbb{H}) are then solved for numerically or in closed form for small n to match the target vertex locations w_k exactly. Confirming that the map is not just boundary-correct but also an honest conformal bijection of the interior \mathbb{H} onto the interior of P — ruling out self-intersection or failure to cover the polygon — is the part left to the Riemann Mapping Theorem's machinery.

Let g(r) = \frac{1}{2\pi r}\oint_{\|\mathbf x-\mathbf a\|=r}u\,ds = \frac{1}{2\pi}\int_0^{2\pi}u(\mathbf a+r(\cos\theta,\sin\theta))\, d\theta. Differentiate under the integral sign (chapter 08) with respect to r: g'(r) = \frac{1}{2\pi}\int_0^{2\pi}\nabla u\cdot(\cos\theta,\sin\theta)\,d\theta = \frac{1}{2\pi r}\oint \nabla u\cdot\hat{\mathbf n}\,ds (recognizing (\cos\theta,\sin\theta) as the outward unit normal). By the Divergence Theorem (chapter 00), this boundary flux equals \frac{1}{2\pi r}\iint_{\text{disk}} \nabla^2u\,dA = 0, since u is harmonic. So g'(r)=0 for every r — g is CONSTANT — and \lim_{r\to0}g(r)=u(\mathbf a) (the average over a shrinking circle approaches the center value by continuity), giving g(r)=u(\mathbf a) for every r.

Suppose u attains an interior maximum at \mathbf a\in D, with u(\mathbf a) = M the largest value u takes anywhere on the closure. By the Mean Value Property, u(\mathbf a) equals the average of u over any small circle centered at \mathbf a. But every value on that circle is \le M (since M is the overall maximum) — an average of values \le M equaling exactly M forces EVERY value on the circle to equal M too (otherwise the average would be strictly less). This holds for circles of every radius small enough to stay in D, so u \equiv M on an entire neighborhood of \mathbf a. The set where u=M is therefore open; it's also closed (as the preimage of a point under a continuous function); on a connected region, a nonempty set that's both open and closed must be the WHOLE region — so u\equiv M everywhere, the constant case. If u isn't constant, no interior maximum can occur, forcing the maximum onto the boundary.

Building the kernel from the Cauchy Integral Formula. Suppose first that u = \operatorname{Re} F for some F analytic on a neighborhood of \overline{\mathbb{D}} (the general continuous-boundary case is recovered afterward by density/limiting arguments standard for this formula, omitted here; the derivation of the kernel itself is the point). Fix z = re^{i\theta} with r<1. The Cauchy Integral Formula applied to F on the circle |\zeta|=1 gives F(z) = \frac{1}{2\pi i}\oint_{|\zeta|=1} \frac{F(\zeta)}{\zeta-z}\,d\zeta. The point z^* = 1/\bar z (the reflection of z across the unit circle) lies outside \overline{\mathbb{D}} when |z|<1, so F(\zeta)/(\zeta - z^*) is analytic on and inside |\zeta|=1, and by Cauchy's Theorem 0 = \frac{1}{2\pi i}\oint_{|\zeta|=1} \frac{F(\zeta)}{\zeta - z^*}\,d\zeta. Subtracting the second identity from the first is how the kernel gets its real-variable form. Since z^* = 1/\bar z, \frac{1}{\zeta-z} - \frac{1}{\zeta-z^*} = \frac{1}{\zeta-z} - \frac{\bar z}{\bar z\zeta - 1} = \frac{1}{\zeta-z} + \frac{\bar z}{1-\bar z\zeta}. On |\zeta|=1, \bar\zeta=1/\zeta, so 1-\bar z\zeta = \zeta(\bar\zeta - \bar z) = \zeta\,\overline{(\zeta-z)}. Putting both terms over the common denominator \zeta(\zeta-z)\overline{(\zeta-z)} = \zeta\,|\zeta-z|^2 and simplifying the numerator, \zeta\,\overline{(\zeta-z)} + \bar z(\zeta-z) = \zeta\bar\zeta - \zeta\bar z + \bar z\zeta - |z|^2 = 1-|z|^2 (using \zeta\bar\zeta=1), so \frac{1}{\zeta-z} - \frac{1}{\zeta-z^*} = \frac{1-|z|^2}{|\zeta-z|^2}\cdot\frac1\zeta. So F(z) = \frac{1}{2\pi i}\oint_{|\zeta|=1} F(\zeta)\left(\frac1{\zeta-z}-\frac1{\zeta-z^*}\right)d\zeta = \frac{1}{2\pi i}\oint_{|\zeta|=1} F(\zeta)\cdot\frac{1-|z|^2}{|\zeta-z|^2}\cdot\frac{d\zeta}{\zeta}. Parametrizing \zeta = e^{i\phi}, d\zeta = i e^{i\phi}d\phi, so d\zeta/\zeta = i\,d\phi, and |\zeta - z|^2 = |e^{i\phi} - re^{i\theta}|^2 = 1 - 2r\cos(\phi-\theta) + r^2 (expand directly), giving F(re^{i\theta}) = \frac{1}{2\pi}\int_0^{2\pi} \frac{1-r^2}{1-2r\cos(\phi-\theta)+r^2}\,F(e^{i\phi})\,d\phi = \frac{1}{2\pi}\int_0^{2\pi} P_r(\theta-\phi)\,F(e^{i\phi})\,d\phi, using \cos(\phi-\theta)=\cos(\theta-\phi). Since P_r(\theta-\phi) is real, taking the real part of both sides (real part commutes with integration against a real weight) gives u(re^{i\theta}) = \operatorname{Re}F(re^{i\theta}) = \frac{1}{2\pi}\int_0^{2\pi}P_r(\theta-\phi)\,\operatorname{Re}F(e^{i\phi})\,d\phi = \frac{1}{2\pi}\int_0^{2\pi}P_r(\theta-\phi)\,u(e^{i\phi})\,d\phi, the Poisson Integral Formula. (Every u harmonic on a disk containing \overline{\mathbb{D}} is locally the real part of some analytic F there, by the harmonic-conjugate construction of harmonic-functions-from-analytic-functions.md applied on a slightly larger simply connected disk, so this covers that case; the boundary regularity needed to extend to merely continuous-on-\overline{\mathbb{D}} data is a standard approximation argument not repeated here.)

Existence and uniqueness for the Dirichlet problem. Given continuous boundary data h, define u by the displayed integral formula with h in place of u(e^{i\phi}). One checks directly that u is harmonic on \mathbb{D} (the Poisson kernel P_r(\theta-\phi), viewed as a function of re^{i\theta}, is itself harmonic — it is, up to the constant real part convention, \operatorname{Re}\frac{e^{i\phi}+z}{e^{i\phi}-z}, the real part of an analytic function of z — and differentiating under the integral sign, justified by uniform convergence on compact subdisks, shows u inherits harmonicity from the kernel) and that u(re^{i\theta}) \to h(e^{i\theta_0}) as re^{i\theta}\to e^{i\theta_0} (the kernel P_r(\theta-\phi) concentrates sharply at \phi=\theta as r\to1^-, acting like an approximate identity, which forces the weighted average to converge to h's value at the limiting boundary point). So this u solves the Dirichlet problem. For uniqueness, if u_1,u_2 both solve it with the same boundary data h, then u_1-u_2 is harmonic on \mathbb{D}, continuous on \overline{\mathbb{D}}, and vanishes on the boundary; the Maximum Principle for harmonic functions (the real-valued analogue of maximum-principle-analytic-functions.md, applied to \pm(u_1-u_2)) forces u_1-u_2\equiv 0 on \overline{\mathbb{D}}, so u_1=u_2.

It suffices to show the two chains agree, element by element, at every parameter value t\in[0,1] where both chains have an element defined at \gamma(t) — the endpoint claim is the case t=1.

Let S = \{t \in [0,1] : \text{the chains' elements at } \gamma(t) \text{ agree on a neighborhood of } \gamma(t)\}. Since f_0=g_0 on D_0 by hypothesis, 0 \in S, so S is nonempty.

S is open. If t_0\in S, the two chains' elements defined near \gamma(t_0) — say f_i on D_i from the first chain and g_j on D_j from the second — agree on some neighborhood U of \gamma(t_0) inside D_i\cap D_j. Both D_i and D_j are open and \gamma is continuous, so for t near t_0, \gamma(t) stays inside D_i\cap D_j (in fact inside U for t close enough), where the two chains' relevant elements still agree — the continuation chains are built so consecutive elements agree on overlaps, and both f_i,g_j continue to represent the respective chains on this stretch until the next transition point. So t is in S too, for t in a neighborhood of t_0: S is open in [0,1].

S is closed. Suppose t_0 is a limit point of S in [0,1]. Let f_i on D_i and g_j on D_j be the two chains' elements defined at \gamma(t_0) (such elements exist and their domains are open neighborhoods of \gamma(t_0), by continuity of \gamma and the finite, overlapping structure of each chain). Since t_0 is a limit point of S, some t_1 \in S has \gamma(t_1) \in D_i\cap D_j (possible for t_1 close enough to t_0, again by continuity and openness), and at t_1 the two chains' elements — which, since D_i, D_j are single elements each valid on their full domains, are again f_i and g_j themselves, restricted appropriately — agree on a neighborhood of \gamma(t_1), by t_1\in S. f_i and g_j are both analytic on D_i \cap D_j (an open set, though possibly disconnected). Restrict attention to the connected component D^* containing \gamma(t_0): it also contains \gamma(t_1), since the arc \gamma([t_1,t_0]) (or [t_0,t_1]) is a connected subset of D_i\cap D_j once t_1 is close enough to t_0, and a connected set meeting D^* lies inside D^*. So f_i and g_j agree on a nonempty open subset of D^*. By the Identity Theorem (identity-theorem-analytic-functions.md), two analytic functions on a connected domain agreeing on a set with a limit point in that domain agree throughout it — an open subset certainly has limit points — so f_i = g_j on all of D^*, in particular on a neighborhood of \gamma(t_0). So t_0 \in S: S is closed.

Conclusion. S is a nonempty, open, and closed subset of the connected set [0,1], so S = [0,1]. In particular 1 \in S, meaning the two chains' final elements agree on a neighborhood of \gamma(1) — in particular f_n(\gamma(1)) = g_m(\gamma(1)), as claimed.

This is stated at the level of rigor this course uses for "simply connected" throughout — as "any closed loop in D can be continuously shrunk to a point without leaving D," equivalently "any two paths in D sharing both endpoints can be continuously slid into one another without leaving D" — rather than the formal homotopy-theoretic machinery a full proof of the general Monodromy Theorem uses; what follows is the idea, not a complete derivation.

Reduce to one pair of paths. Let \gamma_0,\gamma_1 be any two paths in D from z_0 to z_1 along which (f_0,D_0) continues; it suffices to show the two continuations agree at z_1, since the general claim (all continuations along all paths agree) follows by comparing every pair this way.

Slide \gamma_0 to \gamma_1 through a continuous family of paths. Because D is simply connected, \gamma_0 and \gamma_1 (sharing both endpoints) can be continuously deformed into one another through a family of paths \gamma_s, s\in[0,1], each running from z_0 to z_1 and each staying inside D, with \gamma_{s=0}=\gamma_0 and \gamma_{s=1}=\gamma_1.

Continuation along nearby paths in the family agrees. Fix s. Continuing (f_0,D_0) along \gamma_s produces a chain of overlapping disks tracking the path; a path \gamma_{s'} for s' close enough to s stays inside this same chain of disks throughout (the disks have positive radius and \gamma_s\mapsto\gamma_{s'} varies continuously), so continuing along \gamma_{s'} can reuse the same chain of function elements — Uniqueness of Analytic Continuation Along a Path (uniqueness-of-analytic-continuation.md) then forces the two continuations (along \gamma_s and along \gamma_{s'}) to agree at z_1, since they are two continuations of the same starting element along paths close enough to share a common chain of disks. This is the step this proof only sketches: making "close enough to share a chain of disks" precise, uniformly across the whole family \gamma_s, is the technical heart of the general theorem and needs a compactness argument (the interval [0,1] being compact) that is not carried out here.

Chain the agreements together. The set of s\in[0,1] for which the continuation along \gamma_s agrees at z_1 with the continuation along \gamma_0 is, by the previous step, open near every one of its points, and it is nonempty (s=0 trivially belongs); the same kind of open-and-closed connectedness argument used throughout this course (see, e.g., the Identity Theorem or the Maximum Modulus Principle) then shows this set is all of [0,1], in particular containing s=1. So the continuation along \gamma_1 agrees at z_1 with the continuation along \gamma_0.

Why simple connectivity is exactly the hypothesis needed. The argument leans on simple connectivity in exactly one place: the existence of the continuously-varying family \gamma_s sliding \gamma_0 to \gamma_1 without leaving D. On a domain that is not simply connected — \mathbb C\setminus\{0\} is the standard example — two paths from z_0 to z_1 that wind around the missing point a different number of times cannot be slid into one another without crossing the puncture, and indeed continuation of a branch of \log z along such paths genuinely disagrees at the far end: this is exactly the failure this theorem's hypothesis rules out, and it is why \log z and \sqrt z have no single-valued analytic continuation all the way around the origin.

For \operatorname{Re}(z)>0, integrate by parts in \Gamma(z+1) = \int_0^\infty t^{z}e^{-t}\,dt, taking U = t^z, dV = e^{-t}dt, so dU = z\,t^{z-1}dt, V=-e^{-t}: \Gamma(z+1) = \Bigl[-t^z e^{-t}\Bigr]_0^\infty + \int_0^\infty z\,t^{z-1}e^{-t}\,dt. The boundary term vanishes at both ends: as t\to\infty, t^z e^{-t}\to 0 because e^{-t} decays faster than any power of t grows (for \operatorname{Re}(z)>0, |t^z e^{-t}| = t^{\operatorname{Re}(z)}e^{-t}\to0); as t\to 0^+, t^z e^{-t}\to 0 because \operatorname{Re}(z)>0 makes t^{\operatorname{Re}(z)}\to0 while e^{-t}\to1 stays bounded. So the boundary term is 0-0=0, and \Gamma(z+1) = z\int_0^\infty t^{z-1}e^{-t}\,dt = z\,\Gamma(z), the functional equation.

The factorial identity follows by induction as shown above, using \Gamma(1)=1 computed directly from the defining integral.

For the extension claim: on the strip -1<\operatorname{Re}(z)\le0, z\neq0, the right-hand side \Gamma(z+1)/z is well defined since \Gamma(z+1) is already defined there (as \operatorname{Re}(z+1)>0) and z\neq0; declaring \Gamma(z) := \Gamma(z+1)/z on this strip is forced by wanting the functional equation to continue to hold, and it agrees with the original integral definition on the overlap \operatorname{Re}(z)>0 of the two definitions' domains by the identity just proved, so the two definitions glue into one analytic function on \operatorname{Re}(z)>-1, z\neq0. As z\to0, \Gamma(z+1)\to\Gamma(1)=1\neq0 while the denominator z\to0, so \Gamma(z+1)/z has a simple pole at z=0 with residue \lim_{z\to0} z\cdot\frac{\Gamma(z+1)}{z} = \Gamma(1) = 1. Iterating — defining \Gamma(z):=\Gamma(z+1)/z successively on -2<\operatorname{Re}(z)\le-1, then -3<\operatorname{Re}(z)\le-2, and so on — extends \Gamma to a function analytic on \mathbb{C}\setminus\{0,-1,-2,\dots\}, with a simple pole at each z=-n (n\ge0 an integer) of residue \frac{(-1)^n}{n!} (by the same limit computation, now dividing by the accumulated product of the other nearby factors z,z+1,\dots,z+n-1 each approaching -n,-(n-1) ,\dots,-1 respectively, contributing (-1)^n n! in the denominator of the residue's reciprocal). This is the meromorphic continuation of \Gamma to the whole plane.

A fully rigorous derivation is beyond what this course develops — the standard routes go through either the Beta function's integral representation together with a trigonometric substitution and a contour-integral evaluation of the resulting real integral, or through the Weierstrass product for 1/\Gamma(z), both of which need machinery (contour integration of a function with infinitely many poles, or infinite-product theory) past this level. What follows is a proof sketch conveying the idea via the first route.

For \operatorname{Re}(p),\operatorname{Re}(q)>0, the Beta function is defined by B(p,q) = \int_0^1 t^{p-1}(1-t)^{q-1}\,dt, and satisfies the identity B(p,q) = \dfrac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)} (a standard fact proved by writing \Gamma(p)\Gamma(q) as a double integral over (0,\infty)^2 and changing to variables t=\frac{x}{x+y}, s=x+y, which separates the double integral into a product of the B(p,q)-integral in t and a \Gamma(p+q)-integral in s). Setting q = 1-p for 0<\operatorname{Re}(p)<1 (so both p,q have positive real part) gives B(p,1-p) = \Gamma(p)\Gamma(1-p) directly, since \Gamma(p+q)=\Gamma(1)=1. So the reflection formula is equivalent, for 0<\operatorname{Re}(p)<1, to the real-variable identity \int_0^1 t^{p-1}(1-t)^{-p}\,dt = \frac{\pi}{\sin(\pi p)}. Substituting t = \frac{s}{1+s} (so 1-t = \frac{1}{1+s}, dt = \frac{ds}{(1+s)^2}, and t ranging over (0,1) corresponds to s ranging over (0,\infty)) turns the left side into \int_0^\infty \left(\frac{s}{1+s}\right)^{p-1}(1+s)^{p}\,\frac{ds}{(1+s)^2} = \int_0^\infty \frac{s^{p-1}}{1+s}\,ds, after combining the powers of (1+s): -(p-1) + p - 2 = -1. So the identity to be checked becomes the classical real integral \int_0^\infty \frac{s^{p-1}}{1+s}\,ds = \frac{\pi}{\sin(\pi p)} \qquad (0<\operatorname{Re}(p)<1), which is evaluated by contour integration: integrating the (multivalued, hence branch-cut) function \zeta^{p-1}/(1+\zeta) around a keyhole contour that runs just above and just below the positive real axis (avoiding the branch point at 0 and the pole at \zeta=-1), picking up the residue at \zeta=-1 on one hand, and, on the other hand, picking up a factor of 1-e^{2\pi i(p-1)} between the "above" and "below" pieces of the contour because \zeta^{p-1} changes by that factor after \zeta circles the branch point once — comparing the two evaluations of the same contour integral and solving for the real integral produces exactly \pi/\sin(\pi p). This last contour computation is the piece requiring machinery (branch cuts of non-integer powers, the keyhole contour, residues at a pole outside the cut) beyond what is developed in this course, so it is cited here rather than carried out in full; the reduction down to it, from the original claim about \Gamma(z)\Gamma(1-z), is the genuine content of this sketch.

Once the identity is established for 0<\operatorname{Re}(p)<1, both sides of \Gamma(z)\Gamma(1-z) = \pi/\sin(\pi z) are meromorphic functions of z agreeing on this vertical strip, hence agree on their common domain of definition — all non-integer z — by the Identity Theorem (identity-theorem-analytic-functions.md), since the strip has interior points, certainly enough to force agreement on the connected set \mathbb{C}\setminus\mathbb{Z} minus the (removable, from this argument's perspective) strip boundaries.