How do I find a power of any complex number, like (1 + i√3)⁶, using De Moivre's theorem?

In one line

Convert the number to polar form \(r(\cos\theta+i\sin\theta)\); then its \(n\)th power is simply \(r^n(\cos n\theta+i\sin n\theta)\).

Explained simply

Multiplying \((1+i\sqrt3)\) by itself six times by hand is slow and error-prone. In polar form the job is easy: raise the length to the power and multiply the angle. It is like a spiral staircase: each step multiplies your distance from the centre by \(r\) and turns you by \(\theta\).

Technically, if \(z=re^{i\theta}\) then \(z^n=r^ne^{in\theta}=r^n(\cos n\theta+i\sin n\theta)\) for any integer \(n\). This follows from De Moivre's theorem. Negative powers work the same way: \(z^{-n}=r^{-n}(\cos n\theta-i\sin n\theta)\). The method also explains patterns, for example why \((1+i)^n\) is real whenever \(n\) is a multiple of 4: the angle \(n\pi/4\) is then a multiple of \(\pi\).

How it works

1. Find \(r\) and the argument \(\theta\) of the base (check the quadrant).
2. Write \(z^n=r^n(\cos n\theta+i\sin n\theta)\).
3. Evaluate \(\cos n\theta\) and \(\sin n\theta\) using standard angles.
4. For expressions like \((a+ib)^n+(a-ib)^n\), note that conjugates have arguments \(\pm\theta\), so the imaginary parts cancel and you get \(2r^n\cos n\theta\).

Key formula

\(z^n=r^n(\cos n\theta+i\sin n\theta)\), \(r=|z|\), \(\theta=\arg z\) (radians), \(n\) integer.

\((a+ib)^n+(a-ib)^n=2r^n\cos n\theta\) and \((a+ib)^n-(a-ib)^n=2ir^n\sin n\theta\).

Quick example

(a) \((1+i\sqrt3)^6\): \(r=2\), \(\theta=\pi/3\). So \(2^6(\cos2\pi+i\sin2\pi)=64\).

(b) Prove \((1+i)^n+(1-i)^n=2^{\frac n2+1}\cos\dfrac{n\pi}{4}\). Here \(1\pm i=\sqrt2\,e^{\pm i\pi/4}\). The sum is \((\sqrt2)^n\left(e^{in\pi/4}+e^{-in\pi/4}\right)=2^{n/2}\cdot2\cos\dfrac{n\pi}4=2^{\frac n2+1}\cos\dfrac{n\pi}4\).

Common mistakes

• Forgetting to raise the modulus: writing \((1+i\sqrt3)^6=\cos2\pi+i\sin2\pi=1\) instead of 64.
• Wrong argument for numbers in quadrants II–IV, which changes the final answer completely.
• Not reducing large angles: \(\cos\frac{20\pi}{3}=\cos\frac{2\pi}{3}\) after removing \(6\pi\).

How MU asks it

Asked as 3-mark Q.1 parts ("Find \((\sqrt3+i)^{8}\)") or 5/6-mark proofs like "Prove that \((1+\cos\theta+i\sin\theta)^n+(1+\cos\theta-i\sin\theta)^n=2^{n+1}\cos^n\frac\theta2\cos\frac{n\theta}2\)". For the last type, use \(1+\cos\theta=2\cos^2\frac\theta2\), \(\sin\theta=2\sin\frac\theta2\cos\frac\theta2\) to take out \(2\cos\frac\theta2\) first. Show the polar form, the theorem, and the final simplification.