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De Moivre's theorem and powers

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The theorem

Multiplying in polar form multiplies the moduli and adds the arguments. Multiplying a number by itself $n$ times therefore raises the modulus to the power $n$ and multiplies the argument by $n$ .
That statement is De Moivre's theorem:

$(r\,\text{cis}\,\theta)^n = r^n\,\text{cis}(n\theta)$

Each part:
- $r^n$ — the modulus is raised to the power
- $n\theta$ — the argument is multiplied by the power
- $n$ — any integer; the theorem holds for negative $n$ too
$(2\,\text{cis}\,\frac{\pi}{6})^3 = 8\,\text{cis}\,\frac{\pi}{2} = 8i$ .

Why it matters

Expanding $(1 + i)^{10}$ in rectangular form means ten rounds of binomial expansion. In polar form it is two lines.
Any question with an exponent above about 3 is telling you to convert to polar form.

The procedure for a power

Convert the number to polar form, being careful about the quadrant.
Raise the modulus to the power.
Multiply the argument by the power.
Reduce the resulting argument into $(-\pi, \pi]$ by adding or subtracting multiples of $2\pi$ .
Convert back to rectangular form if the question asked for $a + bi$ .
Worked through for $(1 + i)^{10}$ :
- $1 + i = \sqrt{2}\,\text{cis}\,\frac{\pi}{4}$
- Modulus: $(\sqrt{2})^{10} = 2^5 = 32$
- Argument: $10 \times \frac{\pi}{4} = \frac{10\pi}{4} = \frac{5\pi}{2}$
- Reduce: $\frac{5\pi}{2} - 2\pi = \frac{\pi}{2}$
- So $(1 + i)^{10} = 32\,\text{cis}\,\frac{\pi}{2} = 32i$

Negative powers

The theorem holds for negative $n$ , which gives reciprocals:

$(r\,\text{cis}\,\theta)^{-n} = \frac{1}{r^n}\,\text{cis}(-n\theta)$

$(2\,\text{cis}\,\frac{\pi}{3})^{-2} = \frac{1}{4}\,\text{cis}\left(-\frac{2\pi}{3}\right)$ .

Using De Moivre to prove trigonometric identities

Expanding $(\cos\theta + i\sin\theta)^n$ two different ways and equating parts produces multiple-angle identities.
For $n = 2$ :
- By De Moivre: $\text{cis}\,2\theta = \cos 2\theta + i\sin 2\theta$
- By expanding: $(\cos\theta + i\sin\theta)^2 = \cos^2\theta + 2i\sin\theta\cos\theta + i^2\sin^2\theta = (\cos^2\theta - \sin^2\theta) + i(2\sin\theta\cos\theta)$
- Equate real parts: $\cos 2\theta = \cos^2\theta - \sin^2\theta$
- Equate imaginary parts: $\sin 2\theta = 2\sin\theta\cos\theta$
The same method with $n = 3$ gives $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ .

Simplifying expressions before applying the theorem

If a power sits over a quotient, use the division rule first, then apply De Moivre once:

$\left(\frac{r_1\,\text{cis}\,\theta_1}{r_2\,\text{cis}\,\theta_2}\right)^n = \left(\frac{r_1}{r_2}\right)^n \text{cis}\left(n(\theta_1 - \theta_2)\right)$

Doing it in that order avoids raising two numbers to a power unnecessarily.

Tips and Tricks

The 2025 report gives the sharpest possible warning about this topic: "Candidates need to ensure they calculate the correct argument when converting the original equation to polar form. If this initial angle is incorrect then consequently so is all the following work." Everything on this page depends on getting the argument right at step one — so sketch the Argand diagram before you convert.

The modulus is raised to the power; the argument is multiplied by it. Swapping those — writing $nr$ and $\theta^n$ — is the classic slip and it makes both parts wrong.

$(\sqrt{2})^{10}$ is $2^5 = 32$ , not $\sqrt{20}$ or $10\sqrt{2}$ . Convert the surd to an index, $2^{1/2}$ , and multiply the exponents.

Reduce the argument at the end. After multiplying by a large $n$ you will often be several full turns out. Subtract $2\pi$ repeatedly until you land inside $(-\pi, \pi]$ , and state the reduction so the marker can see it.

If a question asks you to prove a multiple-angle identity, the expected method is De Moivre plus equating real and imaginary parts. The report notes candidates "misunderstood what a proof requires" — a proof is a chain of equalities from one side to the other, not a check that the two sides agree at one value of $\theta$ .

Worked ExampleA power in rectangular form

Evaluate $(-\sqrt{3} + i)^6$ , giving your answer in the form $a + bi$ .

Step 1 — Convert to polar form

$a = -\sqrt{3}$ (negative) and $b = 1$ (positive), so the point is in the second quadrant.

Modulus:

$r = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2$

Reference angle, using magnitudes:

$\alpha = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$

Adjust for the second quadrant, where $\theta = \pi - \alpha$ :

$\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$

$-\sqrt{3} + i = 2\,\text{cis}\,\frac{5\pi}{6}$

Step 2 — Apply De Moivre's theorem

Raise the modulus to the power 6 and multiply the argument by 6:

$(2\,\text{cis}\,\tfrac{5\pi}{6})^6 = 2^6\,\text{cis}\left(6 \times \frac{5\pi}{6}\right) = 64\,\text{cis}(5\pi)$

Step 3 — Reduce the argument

$5\pi$ is well outside $(-\pi, \pi]$ . Subtract $2\pi$ twice:

$5\pi - 2\pi = 3\pi, \qquad 3\pi - 2\pi = \pi$

$= 64\,\text{cis}\,\pi$

Step 4 — Convert back to rectangular form

$\cos\pi = -1$ and $\sin\pi = 0$ :

$64(\cos\pi + i\sin\pi) = 64(-1 + 0i) = -64$

Step 5 — Sense-check

The modulus of the answer should be $|{-\sqrt{3} + i}|^6 = 2^6 = 64$ , and $|-64| = 64$ ✓ The answer is purely real, which is what a total argument of a whole multiple of $\pi$ predicts.