Math Core

Complex numbers turn rotations into multiplication. That single fact makes huge powers easy to compute, turns regular polygons into polynomial equations, and lets you "filter" out every third or fourth term of a sum. AIME problems use all three ideas, often dressed up as problems about something else.

Polar form and De Moivre's theorem

Every nonzero complex number can be written as

z = r(\cos\theta + i\sin\theta) = r\operatorname{cis}\theta,

where $r = |z|$ is the distance from $0$ and $\theta$ is the angle from the positive real axis. When you multiply, lengths multiply and angles add:

r_1\operatorname{cis}\theta_1 \cdot r_2\operatorname{cis}\theta_2 = r_1r_2\operatorname{cis}(\theta_1 + \theta_2).

Repeating this gives De Moivre's theorem: $(r\operatorname{cis}\theta)^n = r^n\operatorname{cis}(n\theta)$ . To raise a complex number to a large power, convert to polar form, multiply the angle by $n$ , and reduce it modulo $360^\circ$ .

Two more facts you will use constantly:

$z\bar z = |z|^2$ . If $|z| = 1$ , then $\dfrac{1}{z} = \bar z$ .
If $z = \operatorname{cis}\theta$ , then $z + \dfrac{1}{z} = 2\cos\theta$ and $z^n + \dfrac{1}{z^n} = 2\cos(n\theta)$ .

Roots of unity

Definition

Roots of unity

The $n$ th roots of unity are the $n$ solutions of $z^n = 1$ :

1, \ \omega, \ \omega^2, \ \dots, \ \omega^{n-1}, \qquad \text{where } \omega = \operatorname{cis}\frac{360^\circ}{n}.

They are the vertices of a regular $n$ -gon inscribed in the unit circle, with one vertex at $1$ .

Because they are exactly the roots of $z^n - 1$ , you get a factorization

z^n - 1 = (z - 1)(z - \omega)(z - \omega^2)\cdots(z - \omega^{n-1}).

Everything else follows from this and Vieta's formulas.

Three workhorse facts

Let $\omega = \operatorname{cis}\dfrac{360^\circ}{n}$ with $n \ge 2$ .

Sum: $1 + \omega + \omega^2 + \dots + \omega^{n-1} = 0$ . More generally $\displaystyle\sum_{k=0}^{n-1} \omega^{jk}$ equals $n$ if $n \mid j$ and $0$ otherwise.
Product: $\displaystyle\prod_{k=0}^{n-1}(c - \omega^k) = c^n - 1$ for any number $c$ . Dividing out $c - 1$ and letting $c \to 1$ gives $\displaystyle\prod_{k=1}^{n-1}(1 - \omega^k) = n$ .
Filter: if $f(x) = \sum a_k x^k$ , then

\sum_{n \mid k} a_k = \frac{1}{n}\sum_{j=0}^{n-1} f(\omega^j).

Fact 1 is Vieta (the $z^{n-1}$ coefficient of $z^n - 1$ is $0$ ), or geometry (the vectors to the vertices of a regular polygon cancel). For $j$ not a multiple of $n$ , the numbers $\omega^{jk}$ are again roots of unity summing to zero. Fact 3 follows from fact 1: in $\sum_j f(\omega^j)$ , the term $a_kx^k$ contributes $a_k\sum_j \omega^{jk}$ , which is $na_k$ when $n \mid k$ and $0$ otherwise.

Fact 2 has a geometric meaning: $|1 - \omega^k|$ is the distance from vertex $1$ to vertex $\omega^k$ . So in a regular $n$ -gon inscribed in a unit circle, the product of the distances from one vertex to all the others is $n$ .

Common mistake

When you use $\prod_{k=0}^{n-1}(c - \omega^k) = c^n - 1$ , the order matters: it is $c - \omega^k$ , not $\omega^k - c$ . Flipping all $n$ factors multiplies by $(-1)^n$ . For odd $n$ , $\prod(c + \omega^k) = \prod -(-c - \omega^k) = -\big((-c)^n - 1\big) = c^n + 1$ .

Worked examples

Worked example: A big power

Find the real part of $(1 + i\sqrt3)^{10}$ .

$|1 + i\sqrt3| = 2$ and the angle is $60^\circ$ , so $(1 + i\sqrt3)^{10} = 2^{10}\operatorname{cis}600^\circ = 1024\operatorname{cis}240^\circ$ . Since $\cos 240^\circ = -\dfrac12$ , the real part is $-512$ .

Worked example: Using z + 1/z

A complex number $z$ satisfies $z + \dfrac1z = 1$ . Find $z^{100} + \dfrac{1}{z^{100}}$ .

Multiply by $z$ : $z^2 - z + 1 = 0$ , so $z = \operatorname{cis}(\pm 60^\circ)$ and $|z| = 1$ . Then $z^{100} + z^{-100} = 2\cos(6000^\circ)$ . Since $6000 = 16 \cdot 360 + 240$ , this is $2\cos 240^\circ = -1$ .

Worked example: The roots of unity filter

Compute $\dbinom90 + \dbinom93 + \dbinom96 + \dbinom99$ .

Use $f(x) = (1 + x)^9$ and $n = 3$ , with $\omega = \operatorname{cis}120^\circ$ :

\text{sum} = \frac{f(1) + f(\omega) + f(\omega^2)}{3}.

$f(1) = 512$ . Since $1 + \omega + \omega^2 = 0$ , we have $1 + \omega = -\omega^2 = \operatorname{cis}60^\circ$ , so $f(\omega) = \operatorname{cis}540^\circ = -1$ . Likewise $f(\omega^2) = \overline{f(\omega)} = -1$ . The sum is $\dfrac{512 - 2}{3} = 170$ . Check: $1 + 84 + 84 + 1 = 170$ .

Worked example: Products over roots of unity

Let $\omega = \operatorname{cis}\dfrac{360^\circ}{5}$ . Find $(3 - \omega)(3 - \omega^2)(3 - \omega^3)(3 - \omega^4)$ .

Including $k = 0$ , the full product is $\prod_{k=0}^{4}(3 - \omega^k) = 3^5 - 1 = 242$ . The $k = 0$ factor is $3 - 1 = 2$ , so the answer is $\dfrac{242}{2} = 121$ .

Tip

If a problem involves a sum or product over the vertices of a regular polygon, place the polygon's center at $0$ and write the vertices as $R\,\omega^k$ . Distances become $|z - R\omega^k|$ , and the facts above do the rest.