De Moivre’s Theorem: For any integer $n$ and real $\theta$:
$$(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$$
Equivalently: $(\text{cis}\,\theta)^n = \text{cis}(n\theta)$.
Base case $n=1$: $(\text{cis}\,\theta)^1 = \text{cis}(1\cdot\theta)$. True.
Inductive step: Assume $(\text{cis}\,\theta)^k = \text{cis}(k\theta)$. Then:
$$(\text{cis}\,\theta)^{k+1} = (\text{cis}\,\theta)^k \cdot \text{cis}\,\theta = \text{cis}(k\theta)\cdot\text{cis}\,\theta = \text{cis}((k+1)\theta). \quad\square$$
For negative $n$: use $(\text{cis}\,\theta)^{-1} = \text{cis}(-\theta)$.
Example 1: Compute $(1+i)^{10}$.
Write $1+i = \sqrt{2}\,\text{cis}\,\dfrac{\pi}{4}$. Then:
$$(1+i)^{10} = (\sqrt{2})^{10}\,\text{cis}\,\frac{10\pi}{4} = 32\,\text{cis}\,\frac{5\pi}{2} = 32\,\text{cis}\,\frac{\pi}{2} = 32i$$
Example 2: Simplify $\left(\dfrac{1+i\sqrt{3}}{2}\right)^{12}$.
$\dfrac{1+i\sqrt{3}}{2} = \text{cis}\,\dfrac{\pi}{3}$, so the expression $= \text{cis}(4\pi) = 1$.
The equation $z^n = w$ has exactly $n$ solutions in $\mathbb{C}$.
Method: Write $w = R\,\text{cis}\,\phi$ (principal argument). The $n$ roots are:
$$z_k = R^{1/n}\,\text{cis}!\left(\frac{\phi + 2\pi k}{n}\right), \quad k = 0, 1, \ldots, n-1$$
The roots have equal modulus $R^{1/n}$ and are equally spaced at angles $\dfrac{2\pi}{n}$ on a circle of radius $R^{1/n}$.
Example 3: Find all cube roots of $8i$.
$8i = 8\,\text{cis}\,\dfrac{\pi}{2}$. Cube roots: $z_k = 2\,\text{cis}!\left(\dfrac{\pi/2 + 2\pi k}{3}\right)$ for $k = 0, 1, 2$.
$$z_0 = 2\,\text{cis}\,\frac{\pi}{6} = \sqrt{3} + i$$
$$z_1 = 2\,\text{cis}\,\frac{5\pi}{6} = -\sqrt{3} + i$$
$$z_2 = 2\,\text{cis}\,\left(-\frac{\pi}{2}\right) = -2i$$
Verification: $(\sqrt{3}+i)^3$: expand or use De Moivre: $2^3\,\text{cis}\,\dfrac{3\pi}{6} = 8\,\text{cis}\,\dfrac{\pi}{2} = 8i$. \checkmark
Example 4: Solve $z^4 = -16$.
$-16 = 16\,\text{cis}\,\pi$. Roots: $z_k = 2\,\text{cis}!\left(\dfrac{\pi + 2\pi k}{4}\right)$, $k=0,1,2,3$.
$$z_0 = 2\,\text{cis}\,\frac{\pi}{4} = \sqrt{2}+\sqrt{2}\,i$$
$$z_1 = 2\,\text{cis}\,\frac{3\pi}{4} = -\sqrt{2}+\sqrt{2}\,i$$
$$z_2 = 2\,\text{cis}\,\left(-\frac{3\pi}{4}\right) = -\sqrt{2}-\sqrt{2}\,i$$
$$z_3 = 2\,\text{cis}\,\left(-\frac{\pi}{4}\right) = \sqrt{2}-\sqrt{2}\,i$$
Note: roots in conjugate pairs (coefficients of $z^4 = -16$ are real).
Expanding $(\text{cis}\,\theta)^3 = \text{cis}(3\theta)$ using the binomial theorem:
$$(\cos\theta + i\sin\theta)^3 = \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta$$
Equating real and imaginary parts:
$$\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta = 4\cos^3\theta - 3\cos\theta$$
$$\sin 3\theta = 3\cos^2\theta\sin\theta - \sin^3\theta = 3\sin\theta - 4\sin^3\theta$$
KEY TAKEAWAY: De Moivre’s theorem unifies the calculation of powers and roots of complex numbers. Roots are equally spaced on a circle in the Argand diagram.
EXAM TIP: When finding $n$th roots, always list all $n$ values of $k$ from $0$ to $n-1$ and express final answers in Cartesian or polar form as requested.
COMMON MISTAKE: Using the wrong principal argument for $w$. Always write $w$ in polar form with $\theta \in (-\pi,\pi]$ before applying the root formula.
VCAA FOCUS: Both finding roots of complex numbers and deriving trigonometric identities via De Moivre appear in VCAA Paper 2 extended-answer questions.
