De Moivre's Theorem and Roots (AI HL)
De Moivre's theorem turns raising a complex number to a power into simple arithmetic on its modulus and argument, instead of repeated expansion by hand. This page covers the theorem itself and the related skill of unwinding it to find the \(n\) roots of a complex number, with worked examples and the mistakes that cost marks. It's part of the broader Complex Numbers topic.
11 questions on this sub-topic.
The two formulas
Covered under IB syllabus reference AHL1.13, which asks you to calculate integer powers of a complex number in polar or exponential form. Finding roots builds directly on the same idea and is excellent practice for the underlying skill, even where a paper wouldn't demand it in isolation.
De Moivre's theorem
\[[r(\cos\theta+i\sin\theta)]^n = r^n(\cos n\theta + i\sin n\theta)\]
Equivalently \((re^{i\theta})^n = r^n e^{in\theta}\) in exponential form. Not listed separately in the booklet - it follows directly from the exponential form and index laws that are.
nth roots
\[z^{1/n}=r^{1/n}\,\mathrm{cis}\!\left(\dfrac{\theta+2k\pi}{n}\right),\ k=0,1,\dots,n-1\]
Not in the formula booklet - derived by reversing De Moivre's theorem. Gives \(n\) distinct roots, evenly spaced around a circle of radius \(r^{1/n}\).
Need the full syllabus wording and formula-booklet reference table? See Complex Numbers.
Worked examples
Given \(z = 2e^{i\pi/5}\), find \(z^5\) and explain what it represents geometrically.
Worked solution
\(z^5 = 2^5 e^{i\cdot 5\pi/5} = 32e^{i\pi} = -32.\) M1
\(z^5 = -32.\) A1
Geometrically, the argument is multiplied by 5 (rotation to \(\pi\)) and the modulus raised to the 5th power, M1
landing on the negative real axis. A1
Use De Moivre's theorem to evaluate \((1 + i)^{8}\).
Worked solution
Step 1: \(1+i\) M1
\(= \sqrt2\,\text{cis}\,\tfrac{\pi}{4}.\) A1
Step 2: \((1+i)^8 = (\sqrt2)^8\,\text{cis}(2\pi)\) M1
\(= 16\,\text{cis}(2\pi) = 16.\) A1
Common mistakes
- Forgetting the \(+2k\pi\) when finding roots. Using just \(\theta\) and dividing by \(n\) only gives you one root - you must add multiples of \(2\pi\) to the argument before dividing, running \(k\) from \(0\) to \(n-1\), to catch every root.
- Mixing degrees and radians mid-calculation. De Moivre's theorem works in whichever unit \(\theta\) is given in, but switching units partway through a multi-step power or root calculation is an easy way to lose accuracy marks - stay in one unit throughout.
- Assuming De Moivre's theorem only applies to positive whole-number powers. It extends to negative integers too: \(z^{-n} = r^{-n}\,\mathrm{cis}(-n\theta)\), which is the modulus reciprocated and the argument negated then scaled.
Ready to practise properly?
11 De Moivre's theorem questions, marked instantly like the real exam.
Quick answers
What is De Moivre's theorem?
For a complex number in polar or exponential form, \([r(\cos\theta+i\sin\theta)]^n = r^n(\cos n\theta+i\sin n\theta)\), or equivalently \((re^{i\theta})^n = r^ne^{in\theta}\). Raise the modulus to the power \(n\) and multiply the argument by \(n\).
How do you find the nth roots of a complex number?
Write \(z\) in polar form, then \(z^{1/n}=r^{1/n}\,\mathrm{cis}\!\left(\dfrac{\theta+2k\pi}{n}\right)\) for \(k=0,1,\dots,n-1\). This gives \(n\) distinct roots evenly spaced around a circle of radius \(r^{1/n}\).