Math Core

Treat each point of the plane as a complex number $z = x + yi$ , and rotations become multiplications. Anything involving equilateral triangles, squares, regular polygons or repeated turns is a strong hint to switch to complex numbers.

Points, distances and rotations

The point $(x, y)$ is the number $z = x + yi$ . Then:

The distance between $z$ and $w$ is $|z - w|$ .
The midpoint of $z$ and $w$ is $\dfrac{z + w}{2}$ .
Every $z$ can be written in polar form $z = r(\cos\theta + i\sin\theta) = re^{i\theta}$ , with $r = |z|$ .

Multiplication rotates and scales

$r_1e^{i\theta_1} \cdot r_2e^{i\theta_2} = r_1r_2\,e^{i(\theta_1 + \theta_2)}.$

So multiplying by $e^{i\theta}$ rotates a point about the origin by $\theta$ counterclockwise. To rotate $z$ about a center $c$ :

$z' = c + (z - c)e^{i\theta}.$

Why. Expand $(\cos\alpha + i\sin\alpha)(\cos\beta + i\sin\beta)$ : the real part is $\cos\alpha\cos\beta - \sin\alpha\sin\beta = \cos(\alpha + \beta)$ and the imaginary part is $\sin\alpha\cos\beta + \cos\alpha\sin\beta = \sin(\alpha + \beta)$ . So moduli multiply and angles add. For rotation about $c$ : shift $c$ to the origin, rotate, shift back.

Special cases you'll use constantly:

Multiplying by $i$ rotates by $90^\circ$ .
Multiplying by $\omega = e^{i\pi/3} = \dfrac12 + \dfrac{\sqrt3}{2}i$ rotates by $60^\circ$ . If $a$ , $b$ , $c$ are the vertices of an equilateral triangle in counterclockwise order, then $c - a = (b - a)e^{i\pi/3}$ .

Square with A = 1 + 2i, B = 7 + 5i, C = 4 + 11i, D = -2 + 8i. Each side is the previous one turned by 90 degrees.Open in grapher →

Worked example: Completing a square

Square $ABCD$ is labeled counterclockwise with $A = 1 + 2i$ and $B = 7 + 5i$ . Find $C$ and $D$ .

The side vector is $B - A = 6 + 3i$ . Turning it $90^\circ$ counterclockwise gives $i(6 + 3i) = -3 + 6i$ . So

$C = B + (-3 + 6i) = 4 + 11i, \qquad D = A + (-3 + 6i) = -2 + 8i.$

Worked example: The third vertex of an equilateral triangle

Find the third vertex $c$ of the equilateral triangle with vertices $0$ and $4$ , above the real axis.

Rotate $4$ about $0$ by $60^\circ$ : $c = 4\left(\dfrac12 + \dfrac{\sqrt3}{2}i\right) = 2 + 2\sqrt3\,i$ .

Roots of unity and regular polygons

The solutions of $z^n = 1$ are $1, \zeta, \zeta^2, \dots, \zeta^{n-1}$ , where $\zeta = e^{2\pi i/n}$ . They are the vertices of a regular $n$ -gon inscribed in the unit circle. Two facts do most of the work:

They sum to zero (for $n \ge 2$ ): $1 + \zeta + \dots + \zeta^{n-1} = \dfrac{\zeta^n - 1}{\zeta - 1} = 0$ .
They factor $z^n - 1$ : $z^n - 1 = (z - 1)(z - \zeta)(z - \zeta^2)\cdots(z - \zeta^{n-1})$ .

Products of distances in a regular polygon

If $P$ is the point $z$ and $A_1, \dots, A_n$ are the vertices $R, R\zeta, \dots, R\zeta^{n-1}$ of a regular $n$ -gon with circumradius $R$ , then

$PA_1 \cdot PA_2 \cdots PA_n = |z^n - R^n|.$

In particular, from one vertex to the other $n - 1$ vertices the product is $nR^{n-1}$ .

Proof. Replace $z$ by $z/R$ in the factorization of $z^n - 1$ and multiply by $R^n$ : $z^n - R^n = \prod_{k}(z - R\zeta^k)$ . Take absolute values. For the vertex case, divide $z^n - R^n = (z - R)\prod_{k=1}^{n-1}(z - R\zeta^k)$ by $z - R$ :

$\prod_{k=1}^{n-1}(z - R\zeta^k) = z^{n-1} + Rz^{n-2} + \dots + R^{n-1}.$

Set $z = R$ : the right side is $nR^{n-1}$ . $\blacksquare$

Worked example: Distances from a vertex

A regular $12$ -gon is inscribed in a circle of radius $1$ . Find the product of the distances from one vertex to the other eleven.

By the key idea, the product is $12 \cdot 1^{11} = 12$ .

Worked example: A spiral walk

A robot starts at the origin facing east. It walks $1$ unit, turns $60^\circ$ left, walks $2$ units, turns $60^\circ$ left, walks $3$ units, and so on, finishing with a walk of $6$ units. How far is it from the origin?

With $\omega = e^{i\pi/3}$ , the $k$ -th walk is $k\omega^{k-1}$ , so the endpoint is $S = \sum_{k=1}^{6} k\omega^{k-1}$ . Then

$S - \omega S = 1 + \omega + \omega^2 + \dots + \omega^5 - 6\omega^6 = 0 - 6 = -6,$

since $\omega^6 = 1$ and the sixth roots of unity sum to $0$ . So $S = \dfrac{-6}{1 - \omega}$ . Since $|1 - \omega| = 1$ (the points $0$ , $1$ , $\omega$ form an equilateral triangle), the distance is $6$ .

Common mistake

Rotation direction matters. Multiplying by $e^{i\theta}$ turns counterclockwise. If a problem says the triangle is "clockwise" or puts the new vertex "below" a segment, use $e^{-i\theta}$ , or you'll get the reflected answer.

Tip

To show a triangle is equilateral, or to build one, write one side as a $60^\circ$ rotation of another. To compare lengths, compare $|z|^2 = z\bar z$ , which avoids square roots.

Practice

Practice 1

Square $ABCD$ is labeled counterclockwise, with $A = 3 - i$ and $B = 8 + i$ written as complex numbers. Find $|C|^2 + |D|^2$ .