Math Core

Complex numbers appear on nearly every AMC 12, and they are a favorite for the last third of the test. The key shift is to stop thinking of $a + bi$ as just a pair of numbers and start thinking of it as a point with a length and an angle. Multiplying complex numbers then becomes multiplying lengths and adding angles, and big powers become easy.

The basics

A complex number is $z = a + bi$ with $i^2 = -1$ . You add and multiply like polynomials, replacing $i^2$ with $-1$ . Powers of $i$ cycle with period $4$ :

i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1.

The conjugate of $z = a + bi$ is $\bar{z} = a - bi$ , and the modulus (absolute value) is $|z| = \sqrt{a^2 + b^2}$ , the distance from $0$ to the point $(a, b)$ . The key identity is

z \bar{z} = a^2 + b^2 = |z|^2.

To divide, multiply the top and bottom by the conjugate of the denominator. Also, $|zw| = |z||w|$ and $\left|\dfrac{z}{w}\right| = \dfrac{|z|}{|w|}$ , so you can find the modulus of a messy product without multiplying it out.

Worked example: Powers of i

Find $i + i^2 + i^3 + \dots + i^{2026}$ .

Every block of four consecutive powers sums to $i - 1 - i + 1 = 0$ . Since $2026 = 4 \cdot 506 + 2$ , only the last two terms survive, and they behave like $i^1 + i^2$ . The sum is $-1 + i$ .

Polar form

A complex number with modulus $r$ and angle $\theta$ (measured from the positive real axis) is

z = r(\cos\theta + i\sin\theta).

De Moivre's Theorem

Multiplying complex numbers multiplies their moduli and adds their angles. So

\big(r(\cos\theta + i\sin\theta)\big)^n = r^n\big(\cos n\theta + i\sin n\theta\big).

Worked example: A big power

Compute $(1 + i)^{10}$ .

$(1 + i)^2 = 1 + 2i + i^2 = 2i$ , so $(1 + i)^{10} = (2i)^5 = 32 i^5 = 32i$ .

In polar form: $1 + i$ has modulus $\sqrt{2}$ and angle $45^\circ$ , so its $10$ th power has modulus $(\sqrt{2})^{10} = 32$ and angle $450^\circ$ , which points the same way as $90^\circ$ . That's $32i$ again.

Roots of unity

The solutions of $z^n = 1$ are the $n$ th roots of unity: $z = \cos\dfrac{2\pi k}{n} + i\sin\dfrac{2\pi k}{n}$ for $k = 0, 1, \dots, n - 1$ . They sit at the vertices of a regular $n$ -gon on the unit circle.

The six 6th roots of unity form a regular hexagon on the unit circle.Open in grapher →

Two facts do most of the work:

The $n$ th roots of unity add to $0$ (for $n \ge 2$ ), because they are the roots of $z^n - 1$ , which has no $z^{n-1}$ term.
If $\omega \ne 1$ satisfies $\omega^n = 1$ , then $1 + \omega + \omega^2 + \dots + \omega^{n-1} = 0$ . For cube roots, $\omega^2 + \omega + 1 = 0$ .

Also, $z^n - 1 = (z - 1)(z - \omega)(z - \omega^2)\cdots(z - \omega^{n-1})$ , where $\omega$ is the root with the smallest positive angle. Plugging a number in for $z$ evaluates products over all the roots at once.

Worked example: Cube roots of unity

Let $\omega$ be a nonreal cube root of $1$ . Compute $(1 - \omega + \omega^2)(1 + \omega - \omega^2)$ .

Use $1 + \omega + \omega^2 = 0$ . Then $1 + \omega^2 = -\omega$ , so $1 - \omega + \omega^2 = -2\omega$ . Similarly $1 + \omega = -\omega^2$ , so $1 + \omega - \omega^2 = -2\omega^2$ . The product is $4\omega^3 = 4$ .

Geometry in the complex plane

$|z - w|$ is the distance between the points $z$ and $w$ . So equations like $|z - 4| = |z|$ describe geometric sets: here, the points equally far from $0$ and $4$ , which is the vertical line $\text{Re}(z) = 2$ .

Worked example: Equidistant point

Find the complex number $z$ with $|z| = |z - 4| = |z - 2i|$ .

$|z| = |z - 4|$ means $z$ is on the perpendicular bisector of $0$ and $4$ : real part $2$ . $|z| = |z - 2i|$ means imaginary part $1$ . So $z = 2 + i$ , the center of the circle through $0$ , $4$ and $2i$ .

Common mistake

$|a + bi| = \sqrt{a^2 + b^2}$ , not $a + b$ and not $\sqrt{a^2 - b^2}$ . And $i^2 = -1$ is easy to drop in the middle of a long expansion; after multiplying, scan for every $i^2$ and replace it.

Tip

When a problem has $z + \dfrac{1}{z}$ with a small value, multiply by $z$ to get a quadratic. For instance, $z + \dfrac{1}{z} = 1$ gives $z^2 - z + 1 = 0$ , and multiplying by $z + 1$ shows $z^3 = -1$ . Now high powers of $z$ cycle.

Practice

Practice 1

What is $(2 + 3i)(4 - i)$ ?

Practice 2

What is $\left| \dfrac{(3 + 4i)^3}{(1 - i)^4} \right|$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

What is $(1 + i)^{20}$ ? (It is a real number.)

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

A complex number $z$ with positive real part satisfies $z^2 = -5 + 12i$ . What is $z$ ?

Practice 5

For how many integers $n$ with $1 \le n \le 100$ is $(1 + i)^n$ a real number?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Let $\omega$ be a nonreal complex number with $\omega^3 = 1$ . What is $(2 + \omega)(2 + \omega^2)$ ?

Practice 7

A complex number $z$ satisfies $z + \dfrac{1}{z} = 1$ . What is $z^{2026} + \dfrac{1}{z^{2026}}$ ?

Practice 8

Let $\omega = \cos\dfrac{2\pi}{7} + i\sin\dfrac{2\pi}{7}$ . What is $(2 - \omega)(2 - \omega^2)(2 - \omega^3)(2 - \omega^4)(2 - \omega^5)(2 - \omega^6)$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Real contest practice

2019 AMC 12A, Problem 21: sums of powers of a complex number on the unit circle; reduce exponents using the period.
2019 AMC 12A Problems: the full contest, for timed practice.