De Moivre's Theorem and Roots (AA HL)
Raising a complex number to a large power by repeated Cartesian multiplication is painful - De Moivre's theorem turns it into a single step once the number is in polar form. The same idea runs in reverse to find every \(n\)th root of a complex number, and extends to deriving trig identities like \(\sin3\theta\) in terms of \(\sin\theta\). It's part of the broader Complex Numbers topic.
14 questions on this sub-topic.
The two formulas
Covered under IB syllabus reference AHL1.14: complex conjugate roots of polynomials with real coefficients occur in conjugate pairs; De Moivre's theorem and its extension to rational exponents; powers and roots of complex numbers.
De Moivre's theorem
\(z^n=r^n\,\text{cis}(n\theta)\)
In the formula booklet. Write \(z\) in polar form first, then raise the modulus to the power \(n\) and multiply the argument by \(n\).
\(n\)th roots of a complex number
\(z^{1/n}=r^{1/n}\,\text{cis}\!\left(\dfrac{\theta+2k\pi}{n}\right),\ k=0,1,\ldots,n-1\)
In the formula booklet. Cycling \(k\) through \(0\) to \(n-1\) generates all \(n\) roots, evenly spaced around a circle of radius \(r^{1/n}.\)
Need the full syllabus wording and formula-booklet reference table? See Complex Numbers. For calculator-specific tips, see the parent topic's GDC guidance.
Worked examples
Use De Moivre's theorem to evaluate \((1 + i)^{8}\), giving the answer in Cartesian form.
Worked solution
\(1 + i = \sqrt2\,\text{cis}\tfrac{\pi}{4}.\) M1 A1
\((1+i)^8 = (\sqrt2)^8\,\text{cis}\!\left(8\cdot\tfrac{\pi}{4}\right)\) M1 \(= 16\,\text{cis}(2\pi)\) A1 \(= 16.\) A1
Find the three cube roots of \(8i\), giving each in the form \(a + bi.\)
Worked solution
\(8i = 8\,\text{cis}\tfrac{\pi}{2}.\) M1 A1
\(2\,\text{cis}\!\left(\tfrac{\pi/2 + 2\pi k}{3}\right),\ k = 0,1,2.\) M1 A1
\(k=0\): \(2\,\text{cis}\tfrac{\pi}{6} = \sqrt3 + i.\) A1 \(k=1\): \(2\,\text{cis}\tfrac{5\pi}{6} = -\sqrt3 + i.\) A1 \(k=2\): \(2\,\text{cis}\tfrac{3\pi}{2} = -2i.\) A1
Let \(w=-1+\sqrt3\,i\). The equation \(z^4=w\) has four roots, each with argument \(\theta\) (in radians) where \(-\pi<\theta\le\pi\).
(a) Find \(|w|\).
(b) Find \(\arg w\), giving your answer in radians in terms of \(\pi\).
(c)(i) Find the smallest value of \(\theta\).
(c)(ii) Find the second smallest value of \(\theta\).
(c)(iii) Find the third smallest value of \(\theta\).
(c)(iv) Find the largest value of \(\theta\).
Worked solution
(a) \(|w|=\sqrt{(-1)^2+(\sqrt3)^2}\) M1
\(=\sqrt4=2.\) A1
(b) \(w\) lies in the second quadrant, so \(\arg w=\pi-\tan^{-1}\!\left(\sqrt3\right)=\pi-\dfrac{\pi}{3}\) M1
\(=\dfrac{2\pi}{3}.\) A1
(c)(i) \(w=2\,\text{cis}\dfrac{2\pi}{3}\Rightarrow z^4=2\,\text{cis}\!\left(\dfrac{2\pi}{3}+2k\pi\right).\) M1
\(\theta=\dfrac{\pi}{6}+\dfrac{k\pi}{2}\) gives \(\dfrac{\pi}{6},\ \dfrac{2\pi}{3},\ \dfrac{7\pi}{6},\ \dfrac{5\pi}{3}\); reducing the last two into \((-\pi,\pi]\) gives \(-\dfrac{5\pi}{6}\) and \(-\dfrac{\pi}{3}\). The smallest is \(-\dfrac{5\pi}{6}.\) A1
(c)(ii) \(\theta=-\dfrac{\pi}{3}.\) A1
(c)(iii) \(\theta=\dfrac{\pi}{6}.\) A1
(c)(iv) \(\theta=\dfrac{2\pi}{3}.\) A1
Common mistakes
- Trying to apply De Moivre in Cartesian form. The theorem only works once \(z\) is written as \(r\,\text{cis}\,\theta\) - convert first, then raise the power.
- Stopping after one root. An \(n\)th-root question always has exactly \(n\) solutions - forgetting to cycle \(k\) through every value from \(0\) to \(n-1\) loses marks even if the method is otherwise correct.
- Losing track of \(2k\pi\) inside the argument. Adding multiples of \(2\pi\) before dividing by \(n\) is what spaces the roots evenly around the circle - dropping this term collapses every root onto the same point.
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Quick answers
What is De Moivre's theorem?
De Moivre's theorem states that for a complex number in polar form, \(z^n=r^n\,\text{cis}(n\theta)\). It lets you raise a complex number to a power by raising the modulus to that power and multiplying the argument by it.
How many nth roots does a complex number have?
Exactly \(n\). They all share the same modulus \(r^{1/n}\) and are evenly spaced around a circle centred at the origin, with arguments \(\dfrac{\theta+2k\pi}{n}\) for \(k=0,1,\ldots,n-1\).