4. De Moivre's Theorem & Roots

Video coming soon3 worked examples

Theory

De Moivre's Theorem gives powers of a complex number in polar form:

\bigl[r(\cos\theta + i\sin\theta)\bigr]^n = r^n\bigl(\cos n\theta + i\sin n\theta\bigr)

It is used to evaluate powers, to find $n$ th roots (which are equally spaced around a circle), and to derive multiple-angle trig identities. To find all $n$ th roots, add $2k\pi$ to the argument before dividing by $n$ .

The Fundamental Theorem of Algebra states that a polynomial of degree $n$ has exactly $n$ roots in $\mathbb{C}$ (counting multiplicity). For a polynomial with real coefficients, any complex roots occur in conjugate pairs.

The Golden Rule: when finding roots, write the argument as $\theta + 2k\pi$ first, then take $k = 0, 1, \dots, n-1$ to capture all $n$ distinct roots.

⚠️ Common Examiner Traps

Only finding one root: you must add $2k\pi$ to the argument to obtain all $n$ roots.
Forgetting conjugate pairs: a real polynomial with root $p + qi$ also has root $p - qi$ .
Applying De Moivre outside polar form: convert to modulus–argument form first.

Worked examples

Example 1

Use De Moivre's Theorem to evaluate $(1 + i)^8$ .

Step 1: Write $1 + i$ in polar form. Here $r = \sqrt{2}$ and $\theta = \frac{\pi}{4}$ :

1 + i = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)

Step 2: Apply De Moivre's Theorem with $n = 8$ :

(1+i)^8 = (\sqrt{2})^8\left(\cos 2\pi + i\sin 2\pi\right) = 16(1 + 0)

Step 3: So $(1+i)^8 = 16$ .

Example 2

Find the three cube roots of $8$ , giving your answers in the form $a + bi$ .

Step 1: Write $8$ in polar form, including the $2k\pi$ :

8 = 8\bigl(\cos(0 + 2k\pi) + i\sin(0 + 2k\pi)\bigr)

Step 2: Take the cube root: modulus $8^{1/3} = 2$ , arguments $\frac{2k\pi}{3}$ for $k = 0, 1, 2$ , giving $0,\ \frac{2\pi}{3},\ \frac{4\pi}{3}$ .

Step 3: Evaluate each root:

2,\qquad 2\left(-\tfrac{1}{2} + \tfrac{\sqrt{3}}{2}i\right) = -1 + i\sqrt{3},\qquad -1 - i\sqrt{3}

Example 3

Find the five fifth roots of unity — that is, solve $z^5 = 1$ — and describe their positions on an Argand diagram.

Step 1: Write $1$ in polar form. Crucially, add on multiples of $2\pi$ , since these give the same number but different roots:

1 = \cos(2k\pi) + i\sin(2k\pi), \qquad k \in \mathbb{Z}

Step 2: Apply De Moivre's Theorem with index $\frac{1}{5}$ :

z = \cos\!\left(\frac{2k\pi}{5}\right) + i\sin\!\left(\frac{2k\pi}{5}\right)

Step 3: Take $k = 0, 1, 2, 3, 4$ — five consecutive values give all five distinct roots, and any further $k$ simply repeats them. Adjusting into the range $-\pi < \theta \le \pi$ :

\theta = 0,\ \pm\frac{2\pi}{5},\ \pm\frac{4\pi}{5}

Step 4: Every root has modulus $1$ , so all five lie on the unit circle, equally spaced by $\frac{2\pi}{5}$ , with one of them at $z = 1$ .

This generalises: the $n$ th roots of unity are $n$ equally spaced points on the unit circle, separated by $\frac{2\pi}{n}$ , always including the point $1$ .