1.14 Polynomial roots, De Moivre's theorem, powers and rootsIB Maths: Analysis and Approaches HL: Revision notes

If a polynomial $p(z)$ has real coefficients and $p(z_0) = 0$, then taking conjugates gives $p(z_0^{*}) = 0$ too, because conjugating a sum or product of numbers conjugates each one and leaves real coefficients unchanged. So non-real roots occur in complex conjugate pairs $a \pm bi$.

Each pair multiplies to a real quadratic factor: $(z - (a+bi))(z - (a-bi)) = (z-a)^{2} + b^{2} = z^{2} - 2az + (a^{2}+b^{2}).$ For example the roots $2 \pm 3i$ give $z^{2} - 4z + 13$.

Consequences: a real cubic always has at least one real root; a real quartic has 0, 2 or 4 real roots (counting repeats).

De Moivre's theorem: for $n \in \mathbb{Z}^{+}$, $(r(\cos\theta + i\sin\theta))^{n} = r^{n}(\cos n\theta + i\sin n\theta), \qquad (re^{i\theta})^{n} = r^{n}e^{in\theta}.$ Raise the modulus to the power $n$ and multiply the argument by $n$.

Proof by induction (examinable): true for $n = 1$; assume $z^{k} = \cos k\theta + i\sin k\theta$; multiply by $\cos\theta + i\sin\theta$, expand, and use the compound angle identities $\cos(A+B)$ and $\sin(A+B)$ to get $\cos(k+1)\theta + i\sin(k+1)\theta$; conclude.

The theorem also holds for negative and rational exponents, and you should be aware that it is true for all $n \in \mathbb{R}$. With a negative power: $w^{-2} = r^{-2}\operatorname{cis}(-2\theta)$.

To find a high power, convert to modulus–argument form first. With $w = 1 + \sqrt{3}i = 2e^{i\pi/3}$: $w^{6} = 64e^{2\pi i} = 64, \quad w^{3} = 8e^{i\pi} = -8.$

$w^{n}$ is real when $n\theta$ is a multiple of $\pi$, positive real when it is a multiple of $2\pi$, and purely imaginary when it is an odd multiple of $\frac{\pi}{2}$.

De Moivre also gives multiple-angle identities: expand $(\cos\theta + i\sin\theta)^{3}$ binomially and compare real parts to get $\cos3\theta = 4\cos^{3}\theta - 3\cos\theta$; compare imaginary parts to get $\sin3\theta = 3\sin\theta - 4\sin^{3}\theta$.

To solve $z^{n} = re^{i\theta}$, write the argument in general form $\theta + 2k\pi$: $z = r^{1/n}e^{i\left(\frac{\theta + 2k\pi}{n}\right)}, \quad k = 0, 1, \ldots, n-1.$ There are exactly $n$ distinct $n$th roots. They all have modulus $r^{1/n}$ and their arguments differ by $\frac{2\pi}{n}$, so in the complex plane they are the vertices of a regular $n$-gon centred at the origin.

The $n$th roots of unity solve $z^{n} = 1$: $1, \omega, \omega^{2}, \ldots, \omega^{n-1}$ where $\omega = e^{2\pi i/n}$. Their sum is 0 (a geometric series, or the sum-of-roots result for $z^n - 1 = 0$).