College Algebra rewrote z=a+bi as r(\cos\theta+i\sin\theta). That made multiplication simpler. It stopped being FOIL-and-simplify and became "multiply the distances, add the angles." This section pulls that polar picture back out. It also gives the picture a second spelling, re^{i\theta}. Today that looks like plain notation. But it is the exact object that Chapter 5 spends a whole section turning into a real function, the complex exponential e^z.
Polar form gives directions to the same spot in two different ways. One way is "go 4 blocks, then turn 30 degrees and walk." The other is "go 3 blocks east and 2 blocks north." Here r is how far you walk from home. And \theta, the Greek letter theta, is which way you face when you start.
Euler's formula says the same distance-and-direction pair can be written as re^{i\theta}. That is a different spelling of the same instructions, not a new place on the map.
The argument is not unique, and that is the one wrinkle worth naming. Facing 30° and facing 390° point you the same way. So "the" angle for a given z is really a whole family of equally correct answers. The symbol \operatorname{Arg}z names the one member of that family this course reports by default.
Writing z=a+bi in polar form means recording two things. One is its distance from the origin, r=|z|=\sqrt{a^2+b^2}. The other is a direction angle \theta with a=r\cos\theta and b=r\sin\theta. Then z=r(\cos\theta+i\sin\theta).
That angle is called the argument, written \arg z=\theta. It is never unique. Adding any whole multiple of 2\pi names the same point. So \theta, \theta+2\pi, and \theta-4\pi all describe the same z. The one value inside the interval (-\pi,\pi] is the standard choice, written \operatorname{Arg}z. This course leans on that convention once Chapter 5 needs a single-valued logarithm.
Euler's formula turns the sum in parentheses into one exponential: r(\cos\theta+i\sin\theta)=re^{i\theta}. So every nonzero complex number has an exponential form z=re^{i\theta} with r>0. That form makes multiplication easy to see. Multiplying two exponentials with the same base adds their exponents: r_1e^{i\theta_1}\cdot r_2e^{i\theta_2}=r_1r_2\,e^{i(\theta_1+\theta_2)}. Moduli multiply and arguments add. No ordinary law of exponents is broken anywhere.
Quick review
Practice from earlier coursework, pulled in for this topic.
From College Algebra
What is \theta, the argument of -3 + 3\sqrt3\,i, in radians, restricted to [0, 2\pi)? Enter an exact expression (e.g. pi/6).
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
The reference angle is \tan^{-1}\!\left(\dfrac{3\sqrt3}{3}\right) = \tan^{-1}(\sqrt3) = \dfrac{\pi}{3}. Since a=-3<0 and b=3\sqrt3>0, this is Quadrant II, so the argument is \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}.
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Why
Find the reference angle from the magnitudes first, then place it by quadrant.
From College Algebra
Convert 5(\cos 210° + i\sin 210°) to rectangular form. What is its real part?
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Why
The real part is a = r\cos\theta = 5\cos 210° = 5\left(-\frac{\sqrt3}{2}\right) = -\frac{5\sqrt3}{2} \approx -4.3301.
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Why
The real part in rectangular form is a = r\cos\theta.
Theorem
Euler's Formula
also known as: e^{i\theta} = \cos\theta + i\sin\theta
Recalled from Calculus II — Integral Calculus & Infinite Series · Euler's formula and complex exponentials — first taught there.
For every real number \theta, e^{i\theta} = \cos\theta + i\sin\theta.
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.
Reading r and \theta off a known picture
Drag the angle. The horizontal and vertical coordinates you already know as \cos\theta and \sin\theta are exactly z's real and imaginary parts when r=1.
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Recipe Convert between rectangular and exponential form Use when: you have z=a+bi and need z=re^{i\theta}, or you have z=re^{i\theta} and need a+bi Goal: the same complex number rewritten in whichever of the two forms you didn't start with, with r>0 and \theta picked correctly by quadrant
Quick recipe
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1
Look at what you're actually given: a rectangular number a+bi, or an exponential number re^{i\theta}. The two directions use different halves of this recipe, so this is the fork that decides where to start.
- given a+bi — continue to Step 2 (find r and \theta).
- given re^{i\theta} — skip ahead to Step 4 (expand with Euler's formula).
Why: the modulus-argument pair and the exponential form are two spellings of the same data, so converting one way is just running the other direction's steps in reverse.
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2
Compute r=|z|=\sqrt{a^2+b^2} directly from the real and imaginary parts.
Example
For z=3-3\sqrt3\,i: a=3, b=-3\sqrt3, so r=\sqrt{3^2+(3\sqrt3)^2}=\sqrt{9+27}=\sqrt{36}=6.
Why: r is just z's distance from the origin, and the Pythagorean theorem gives that distance straight from the horizontal and vertical legs a and b.
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3
Compute the reference angle from \tan(\text{ref})=|b/a|, then adjust its sign, and if needed subtract it from \pi, so the result lands in (-\pi,\pi] and matches the quadrant z actually sits in.
Example
Continuing with z=3-3\sqrt3\,i: a=3>0, so this is the first case. \tan(\text{ref})=\dfrac{3\sqrt3}{3}=\sqrt3, so the reference angle is \pi/3. Since b=-3\sqrt3<0, \theta=-\pi/3.
- a>0 (quadrant I or IV) — \theta is the reference angle itself, positive if b>0 and negative if b<0.
- a<0 (quadrant II or III) — \theta is \pi minus the reference angle if b>0, or the reference angle minus \pi if b<0.
- a=0 — skip the reference-angle triangle entirely; \theta=\pi/2 if b>0 and \theta=-\pi/2 if b<0.
⚡ Shortcut
When a=0 or b=0, z sits right on an axis and \theta is one of 0,\pi/2,\pi,-\pi/2, read off directly with no reference-angle triangle needed at all.
Why: the reference angle alone only ever gives an acute angle; the quadrant is what tells you which of the four sign/offset combinations actually matches where z sits.
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4
Going the other direction, start from z=re^{i\theta} and rewrite the exponential as \cos\theta+i\sin\theta using Euler's formula, still carrying the factor of r out front.
Example
For z=4e^{-i\pi/3}: z=4\Big(\cos(-\pi/3)+i\sin(-\pi/3)\Big).
Why: Euler's formula is exactly the identity that lets \cos\theta+i\sin\theta and e^{i\theta} stand in for each other; nothing else in this step is new.
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5
Look up (or compute) \cos\theta and \sin\theta, then multiply both by r to land in a+bi form.
Example
\cos(-\pi/3)=\tfrac12 and \sin(-\pi/3)=-\tfrac{\sqrt3}{2}, so z=4\left(\frac12-\frac{\sqrt3}{2}i\right)=2-2\sqrt3\,i.
Why: distributing r across both the cosine term and the sine term is what turns the compact exponential form back into an explicit real part and imaginary part.
Common mistakes
- Forgetting the reference angle is always positive and acute, and needs a quadrant adjustment tacked on afterward.
- Reporting an angle outside (-\pi,\pi] instead of adding or subtracting a full 2\pi to land in the principal range.
- Reading r straight off a Pythagorean sum without taking the positive square root.
- Multiplying only one of \cos\theta and \sin\theta by r instead of both, when expanding re^{i\theta}.
Worked example
Converting a rectangular number to exponential form
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1
Write z=-2+2\sqrt3\,i in the form re^{i\theta}, the same recipe as above: Step 2 for r, then Step 3 for \theta. r = \sqrt{(-2)^2+(2\sqrt3)^2} = \sqrt{4+12} = 4
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2
The reference angle satisfies \tan(\text{ref})=\dfrac{2\sqrt3}{2}=\sqrt3, so the reference angle is \pi/3. The point sits in the second quadrant (negative real part, positive imaginary part), so \theta = \pi-\frac{\pi}{3} = \frac{2\pi}{3}
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3\boxed{z = 4e^{i2\pi/3}}
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Worked example
Evaluating an exponential form back in a+bi
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1
Evaluate e^{i5\pi/6} in a+bi form.
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2
By Euler's formula, e^{i5\pi/6}=\cos(5\pi/6)+i\sin(5\pi/6).
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3
\cos(5\pi/6)=-\tfrac{\sqrt3}{2} and \sin(5\pi/6)=\tfrac12, both read straight off the unit circle above. \boxed{e^{i5\pi/6} = -\frac{\sqrt3}{2}+\frac12 i}
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Where this goes next
The launchpad for Chapter 5's exponential function
Right now e^{i\theta} is only a compact way to write \cos\theta+i\sin\theta. It is a notation. It is not yet a function you could feed an arbitrary complex number. Chapter 5 defines e^{z} for every z=x+iy at once. It then proves that e^{i\theta} is what this reduces to when x=0. Everything reviewed here about magnitudes multiplying and angles adding survives unchanged in that fuller definition.
Check your understanding (counts toward your progress)
Find the modulus of the product \left(2e^{i\pi/6}\right)\left(3e^{i\pi/3}\right).
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Why
2\cdot3=6.
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Why
Moduli of an exponential-form product simply multiply.
Find the argument of that same product, \left(2e^{i\pi/6}\right)\left(3e^{i\pi/3}\right), in terms of \pi.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
\pi/6+\pi/3=\pi/6+2\pi/6=3\pi/6=\pi/2.
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Why
Arguments of an exponential-form product simply add.
How does \arg z differ from \operatorname{Arg}z?
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Neither is restricted to being positive by definition. \operatorname{Arg}z can be negative, for instance for a point below the real axis.
\arg z is genuinely multivalued — \operatorname{Arg}z singles out just one of its infinitely many values.
\operatorname{Arg}z is defined for any nonzero complex number, real or not — it isn't restricted to real inputs.
Why
Adding any multiple of 2\pi to a valid argument gives another valid argument, so \arg z names the whole family; \operatorname{Arg}z is the single representative chosen to lie in (-\pi,\pi].
Adding a full turn, 2\pi, to an angle points in the exact same direction.
Find \operatorname{Arg}(z) for z=-4, a negative real number. Enter your answer in terms of \pi (using * for multiplication).
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
-4 points in the exact opposite direction from the positive real axis, an angle of \pi.
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Why
A negative real number sits directly on the negative real axis.
Practice: Converting between rectangular and exponential form (5 problems · doesn’t affect your progress)
Converting a complex number from rectangular to polar form means finding...
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\theta alone doesn't pin down a specific point — the same angle describes infinitely many distances from the origin.
r alone doesn't pin down a specific point either — the same distance describes infinitely many directions.
a and b are exactly the rectangular coordinates already given — polar form asks for a different pair, r and \theta.
Why
Polar form records both the distance r from the origin and the direction angle \theta.
r and \theta
What is the modulus r of z=3i?
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Why
r=\sqrt{0^2+3^2}=3.
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Why
Find \theta for z=3i, in terms of \pi.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
A point straight up the imaginary axis has argument \pi/2.
Correct answer
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Why
z=3i sits straight up on the positive imaginary axis.
Convert z=-1-i to exponential form re^{i\theta}. Enter r.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
r=\sqrt{1+1}=\sqrt2.
Correct answer
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Why
r=\sqrt{(-1)^2+(-1)^2}.
A student says the argument of z=-5 (a negative real number) is 0, since it lies on the real axis. What's the mistake?
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Every nonzero complex number, negative reals included, has a well-defined argument.
Degrees versus radians doesn't change which direction -5 points — the actual issue is which angle was picked.
-5 points in the exact opposite direction from angle 0 — that's angle \pi, not 0.
Why
The real axis has two directions: angle 0 for positive reals, angle \pi for negative reals. -5 has argument \pi.
Negative reals lie on the axis at angle \pi, not 0 — angle 0 is for POSITIVE reals