The last section showed that multiplying two complex numbers in exponential form multiplies their distances and adds their angles. De Moivre's theorem applies that fact to a number times itself, over and over. It also gives you, almost for free, the one clean way to undo a power. That means finding every complex number that raises to a given result.
Watch what raising to a power does to a point in the plane. It is the first sighting of an idea that Chapter 3 promotes to the definition of the derivative itself.
Raising a number to a power is rotate-and-scale, repeated. Squaring a number at angle 30° puts the result at angle 60°, with the distance squared.
An nth root undoes exactly that. So it has to land somewhere that reaches the target after n repeats of rotate-and-scale. More than one starting angle works, because rotating by one full circle changes nothing about where you land. That is why the n roots come out evenly spaced, like n slices of one pie. Each slice, taken n times over, ends at the same final angle.
Here is a habit this breaks. For real numbers, "the square root of 4" quietly means the positive one. But a complex nth root is never a single answer. It is n different, equally correct numbers. A problem that says "find the cube root" with no more context is asking the wrong question.
De Moivre's theorem is the exponential-form product rule applied to one number multiplied by itself n times: [re^{i\theta}]^n=r^ne^{in\theta}. So raising to the nth power multiplies the angle by n. It also raises the modulus to the nth power.
Now run that backward to solve z^n=w for a nonzero w=Re^{i\varphi}. Every solution must have modulus R^{1/n}, the ordinary positive real nth root. Its angle must satisfy n\theta=\varphi+2\pi k for some integer k. That is because angles that differ by a whole turn describe the same w. This gives exactly n distinct solutions: z_k = R^{1/n}\left(\cos\frac{\varphi+2\pi k}{n}+i\sin\frac{\varphi+2\pi k}{n}\right), \qquad k=0,1,\dots,n-1. They are spaced 2\pi/n apart around a circle of radius R^{1/n}. Taking k=n would repeat z_0, since that angle differs from z_0's by a full 2\pi.
The special case w=1 gives the nth roots of unity. They are evenly spaced around the unit circle, and z_0=1 is always one of them.
Quick review
Practice from earlier coursework, pulled in for this topic.
From College Algebra
z_1 has modulus 7 and z_2 has modulus 4. What is the modulus of z_1z_2?
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Why
Multiplying complex numbers multiplies their moduli: 7 \times 4 = 28.
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From College Algebra
z lies on the unit circle (modulus 1) with argument 80°. What is the modulus of z^{15}?
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Why
De Moivre's Theorem raises the modulus to the nth power: 1^{15}=1 — a number on the unit circle never leaves it, no matter the power.
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Quick review
Practice from earlier coursework, pulled in for this topic.
From College Algebra
A complex number has modulus 1 (it lies on the unit circle). What is the modulus of each of its nth roots, for any positive integer n?
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Why
1^{1/n}=1 for every positive integer n — every root of a number on the unit circle also lands on the unit circle.
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From College Algebra
A complex number z has argument \frac{\pi}{2}. Its cube roots have angles \dfrac{\frac{\pi}{2}+2\pi k}{3} for k=0,1,2. Which value of k gives the root with angle \frac{5\pi}{6}?
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Why
Set \dfrac{\frac{\pi}{2}+2\pi k}{3}=\frac{5\pi}{6}. Multiply both sides by 3: \frac{\pi}{2}+2\pi k=\frac{5\pi}{2}. Subtract \frac{\pi}{2}: 2\pi k=2\pi, so k=1.
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Why
Set the angle formula equal to \frac{5\pi}{6} and solve for k.
Theorem
De Moivre's Theorem
Recalled from College Algebra · Products, quotients, and De Moivre's Theorem — first taught there.
For any positive integer n, \big[r(\cos\theta+i\sin\theta)\big]^n = r^n\big(\cos(n\theta)+i\sin(n\theta)\big).
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
Multiplying by i is a quarter-turn
Drag the point at 1+i — multiplying it by i (fixed at angle 90°) always rotates it a quarter turn without changing its distance from the origin.
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Worked example
A power by De Moivre's theorem
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1
Find (1+i)^6. |1+i| = \sqrt{2}, \qquad \arg(1+i) = \frac{\pi}{4}, \qquad \text{so } 1+i=\sqrt2\, e^{i\pi/4}
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2(1+i)^6 = \left(\sqrt2\right)^6 e^{i\cdot 6\pi/4} = 8\, e^{i3\pi/2}(\sqrt2)^6 = 2^3 = 8
(\sqrt2)^6 = 2^3 = 8
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3
\cos(3\pi/2)=0 and \sin(3\pi/2)=-1, so \boxed{(1+i)^6 = 8(0-i) = -8i}
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Worked example
The three cube roots of -8i
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1
Find every z with z^3=-8i. -8i has modulus 8 and argument -\pi/2, so z^3=8e^{i(-\pi/2)}. Each root has modulus 8^{1/3}=2 and angle \dfrac{-\pi/2+2\pi k}{3}.
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2
k=0: angle -\pi/6, giving z_0=2\!\left(\cos(-\tfrac{\pi}{6})+i\sin(-\tfrac{\pi}{6})\right)=\sqrt3-i.
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3
k=1: angle \pi/2, giving z_1=2(\cos\tfrac{\pi}{2}+i\sin\tfrac{\pi}{2})=2i. \quad k=2: angle 7\pi/6, giving z_2=2(\cos\tfrac{7\pi}{6}+i\sin\tfrac{7\pi}{6})=-\sqrt3-i.
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4
The three roots sit 2\pi/3 apart around the circle of radius 2, and z_0+z_1+z_2=(\sqrt3-\sqrt3)+(-i+2i-i)=0 — a useful check, since the roots of z^3-(-8i)=0 must sum to 0 (no z^2 term). The figure shows this spacing directly: multiplying z_0 by \cos(2\pi/3)+i\sin(2\pi/3), the unit rotation by exactly one root's worth of angle, lands squarely on z_1 — the same 2\pi/3 turn that would carry z_1 to z_2 and z_2 back around to z_0. \boxed{z=\sqrt3-i,\ \ 2i,\ \ -\sqrt3-i}
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Where this goes next
A first sighting of what differentiating in C will mean
Multiplying by a fixed complex number is always exactly a rotation combined with a scaling. No other outcome is possible. Chapter 3 proves something about functions of a complex variable. Wherever such a function has a derivative, it treats points very close together in this same way. That single fact about shapes makes complex differentiability a far stronger condition than its real-variable namesake. It is not just a two-dimensional copy of it.
Check your understanding (counts toward your progress)
Use De Moivre's theorem to find (\sqrt3+i)^4. Enter your answer as a complex number.
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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
(\sqrt3+i)^4=2^4e^{i4\pi/6}=16e^{i2\pi/3}=16\left(-\tfrac12+\tfrac{\sqrt3}{2}i\right)=-8+8\sqrt3\,i.
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Why
First write \sqrt3+i in exponential form: r=2, \theta=\pi/6.
A student computes (1+i)^2 using De Moivre and gets 2e^{i\pi/2}, then rewrites this as 2i. Is this correct?
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Squaring multiplies the modulus by itself: (\sqrt2)^2=2, not \sqrt2\cdot2=4 — there's no reason to double it instead of squaring it.
The angle DOES double: \pi/4 to \pi/2 is exactly doubling. \pi would be one power too many.
2i is exactly 0+2i, a perfectly ordinary complex number with a=0 — a zero real part isn't invalid.
Why
1+i=\sqrt2\,e^{i\pi/4}, so (1+i)^2=2e^{i\pi/2}=2i. The computation is correct.
Convert 1+i to exponential form first, then apply De Moivre's theorem for n=2.
How many distinct 6th roots does a nonzero complex number have?
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Why
An nth root equation z^n=w (for w\ne 0) always has exactly n distinct solutions.
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For z with r=2, \theta=\pi/3, find |z^5|.
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Why
|z^5|=r^5=2^5=32.
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Why
Moduli raised to the nth power just get raised to that power themselves.
Practice: Powers and roots, De Moivre's way (5 problems · doesn’t affect your progress)
To raise a complex number to a power efficiently, the first step is...
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This isn't how complex-number exponentiation works at all — a+bi raised to a power isn't na+nbi.
Squaring the modulus only handles the SECOND power — De Moivre's theorem is needed for any other exponent.
Both the modulus and the argument matter — dropping the modulus loses the resulting size entirely.
Why
Once in re^{i\theta} form, raising to the nth power is immediate: r^ne^{in\theta}.
Convert it to polar or exponential form
Find (2i)^3. Enter your answer as a complex number.
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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
(2i)^3=8i^3=8(-i)=-8i.
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(2i)^3=8i^3, and i^3=-i.
Find (1-i)^4 using De Moivre's theorem.
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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
(1-i)^4=(\sqrt2)^4e^{-i\pi}=4(\cos(-\pi)+i\sin(-\pi))=4(-1+0i)=-4.
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1-i=\sqrt2\,e^{-i\pi/4}.
Find one cube root of 27 other than 3 itself.
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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
At angle 2\pi/3: 3\left(\cos(2\pi/3)+i\sin(2\pi/3)\right)=3\left(-\tfrac12+\tfrac{\sqrt3}{2}i\right)=-\tfrac32+\tfrac{3\sqrt3}{2}i.
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Why
27=27e^{i0}; the cube roots have modulus 3 and angles 0,2\pi/3,4\pi/3.
A student says z^4=16 has exactly one solution, z=2, since 2^4=16. What's wrong?
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2^4=16 is correct arithmetic — the issue is elsewhere.
Every real number is a complex number (z=2+0i) — that's not the problem.
z=2 IS one of the four roots — the mistake is stopping there instead of finding the other three.
Why
The other three 4th roots of 16 are 2i, -2, and -2i. A real-number habit does not survive the move to \mathbb C. That habit is assuming one obvious answer is the only one.
16 has four 4th roots total, not just the positive real one