Polar Form and De Moivre's Theorem

1. Motivation

Rectangular form is powerful but clumsy for multiplication. Compare:

$\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)^{12}$

Expanding this directly requires twelve rounds of FOIL. In polar form, it is $\left(e^{i\pi/6}\right)^{12} = e^{2\pi i} = 1$. Done in one line.

The central insight: multiplication in polar form is rotation and scaling.

The Complex Plane in Polar Coordinates

A point $z = a + bi$ sits at distance $r = |z|$ from the origin. The angle from the positive real axis, measured counterclockwise, is $\theta = \arg(z)$. These two quantities, the modulus and argument, completely determine $z$.

2. Polar Form

For any nonzero $z = a + bi$, define:

$|z| = r = \sqrt{a^2 + b^2}, \qquad \arg(z) = \theta \text{ where } \tan\theta = \frac{b}{a}$

The principal argument $\text{Arg}(z) \in (-\pi, \pi]$ is the canonical choice. The full argument is $\theta + 2\pi k$ for any integer $k$.

Two equivalent notations for polar form:

$z = r(\cos\theta + i\sin\theta) \qquad \text{(trigonometric form)}$

$z = re^{i\theta} \qquad \text{(exponential form, via Euler's formula)}$

Euler's formula: $e^{i\theta} = \cos\theta + i\sin\theta$. This follows from the Taylor series of $e^x$, $\cos x$, and $\sin x$.

Worked conversions.

Rectangular	$r$	$\theta$	Polar
$1 + i$	$\sqrt{2}$	$\pi/4$	$\sqrt{2}\,e^{i\pi/4}$
$-1 + \sqrt{3}i$	$2$	$2\pi/3$	$2e^{2i\pi/3}$
$-i$	$1$	$-\pi/2$	$e^{-i\pi/2}$
$-3$	$3$	$\pi$	$3e^{i\pi}$

3. Geometry of Multiplication

For $z_1 = r_1 e^{i\alpha}$ and $z_2 = r_2 e^{i\beta}$:

$z_1 z_2 = r_1 r_2\, e^{i(\alpha + \beta)}$

Magnitudes multiply. Arguments add.

Multiplication as Rotation and Scaling

Multiplying $z_1$ by $z_2$ rotates $z_1$ counterclockwise by $\arg(z_2)$ and scales it by $|z_2|$. When $|z_2| = 1$, this is a pure rotation with no change in length.

The rule for division is the natural inverse:

$\frac{z_1}{z_2} = \frac{r_1}{r_2}\,e^{i(\alpha - \beta)}$

Dividing by $z_2$ rotates clockwise by $\arg(z_2)$ and scales by $1/|z_2|$.

Key principle: Multiplying any complex number by a unit complex number $e^{i\theta}$ rotates it by $\theta$ about the origin. To rotate a point $z$ about a center $c$ by angle $\theta$, compute:

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

4. Rotations

Rotating by $90^\circ$

Multiplication by $i = e^{i\pi/2}$ rotates counterclockwise by $90^\circ$. In coordinates, $(a + bi) \cdot i = -b + ai$: the old imaginary part becomes the new real part (negated), and the old real part becomes the new imaginary part.

Rotating by $180^\circ$

Multiplication by $-1 = e^{i\pi}$ maps $z \mapsto -z$, a $180^\circ$ rotation about the origin.

Rotating by an arbitrary angle $\theta$

Multiply by $e^{i\theta} = \cos\theta + i\sin\theta$.

Example. Rotate $z = 3 + i$ by $\pi/3$ about the origin.

$z' = e^{i\pi/3}(3+i) = \left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)(3+i) = \frac{3-\sqrt{3}}{2} + \frac{3\sqrt{3}+1}{2}i$

Shortcut. To check that triangle $ABC$ is equilateral, verify:

$(C - A) = e^{\pm i\pi/3}(B - A)$

This converts an equilateral triangle condition into a single complex multiplication.

Rotation Characterization of Equilateral Triangles

The triangle $ABC$ is equilateral if and only if $C - A = e^{i\pi/3}(B-A)$ or $C - A = e^{-i\pi/3}(B-A)$. Two orientations, one formula.

5. De Moivre's Theorem

Theorem. For any integer $n$:

$(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$

Proof. Write $\cos\theta + i\sin\theta = e^{i\theta}$ via Euler's formula. The exponential law gives $(e^{i\theta})^n = e^{in\theta}$, and another application of Euler's formula yields $e^{in\theta} = \cos(n\theta) + i\sin(n\theta)$. For negative $n$, observe $(e^{i\theta})^{-1} = e^{-i\theta}$, so the same argument applies.

Trig identities from De Moivre. Expand $(\cos\theta + i\sin\theta)^3$ via binomial theorem and match real and imaginary parts:

$\cos(3\theta) = \cos^3\theta - 3\cos\theta\sin^2\theta, \qquad \sin(3\theta) = 3\cos^2\theta\sin\theta - \sin^3\theta$

This method generates all multiple-angle formulas mechanically.

Powers of a Unit Complex Number

Repeated multiplication by $e^{i\theta}$ steps around the unit circle in equal angular increments. After $n$ steps, you land at $e^{in\theta}$: this is De Moivre's theorem made visual.

6. De Moivre Bashing

Computing powers quickly

Write the number in polar form, apply De Moivre, convert back.

Example. Compute $(1+i)^{20}$.

$1 + i = \sqrt{2}\,e^{i\pi/4}$

$(1+i)^{20} = \left(\sqrt{2}\right)^{20} e^{i \cdot 20\pi/4} = 2^{10} e^{5\pi i} = 1024 \cdot e^{i\pi} = -1024$

Extracting trig identities

To find $\cos^5\theta$ in terms of multiple angles, write $\cos\theta = \frac{e^{i\theta}+e^{-i\theta}}{2}$ and expand:

$\cos^5\theta = \frac{1}{32}\left(e^{5i\theta} + 5e^{3i\theta} + 10e^{i\theta} + 10e^{-i\theta} + 5e^{-3i\theta} + e^{-5i\theta}\right)$

$= \frac{1}{16}\left(\cos 5\theta + 5\cos 3\theta + 10\cos\theta\right)$

This technique, expressing a power of cosine as a linear combination of cosines of multiples, appears frequently on the AIME.

Finding real and imaginary parts

Example. Find $\text{Re}\left[(2+2i)^8\right]$.

$(2+2i) = 2\sqrt{2}\,e^{i\pi/4}$, so $(2+2i)^8 = (2\sqrt{2})^8 e^{2\pi i} = 2^{12} = 4096$. Real part is $4096$.

Detecting roots of unity

If $\omega = e^{2\pi i/n}$, then $\omega^n = 1$ and $1 + \omega + \omega^2 + \cdots + \omega^{n-1} = 0$. This identity is the key to the roots-of-unity filter technique.

7. Roots of Complex Numbers

To solve $z^n = w$ where $w = \rho e^{i\phi}$:

$z = \rho^{1/n}\, e^{i(\phi + 2\pi k)/n}, \qquad k = 0, 1, \ldots, n-1$

There are exactly $n$ roots, evenly spaced at angles $2\pi/n$ apart on a circle of radius $\rho^{1/n}$.

The nth Roots of Unity

The $n$th roots of unity form a regular $n$-gon inscribed in the unit circle. Their sum is always zero. This symmetry is exploited constantly in problems.

Extracting Cube Roots Geometrically

To find the cube roots of $w = 8e^{i\pi/3}$: take the cube root of the modulus to get radius $2$, divide the argument by $3$ to get the first root at angle $\pi/9$, then add $2\pi/3$ twice to get the other two. All three roots lie on a circle of radius $2$.

Key fact. The roots of $z^n = 1$ are $e^{2\pi i k/n}$ for $k = 0, \ldots, n-1$. These are the vertices of a regular $n$-gon centered at the origin with one vertex at $1$.

8. Common Olympiad Techniques

Choosing the Right Technique

The right technique depends on the structure of the problem. Multiplication, powers, and angles call for polar form. Sums and real-part extraction call for rectangular. Geometric arguments about angles and distances call for modulus-argument reasoning.

Technique 1: Convert to polar immediately. Any problem involving products, quotients, or powers of complex numbers should be converted to polar form before doing any algebra.

Technique 2: Rotate instead of coordinate-bashing. When a problem asks about a geometric transformation, represent it as multiplication by $e^{i\theta}$ rather than using rotation matrices or coordinates.

Technique 3: Use unit complex numbers for angles. If $|z| = 1$, then $z = e^{i\theta}$ for some $\theta$, and $\bar{z} = e^{-i\theta} = z^{-1}$. This simplifies conjugate expressions dramatically.

Technique 4: Exploit symmetry. If $f(z)$ is symmetric under rotation by $2\pi/n$, the answer likely involves $n$th roots of unity. Summing over all rotations and using $\sum_{k=0}^{n-1}\omega^{jk} = 0$ for $j \not\equiv 0\pmod{n}$ filters out all but the symmetric terms.

Technique 5: Use conjugates cleverly. $z + \bar{z} = 2\,\text{Re}(z)$ and $z\bar{z} = |z|^2$. When you see a real-part extraction in a sum, write the sum twice with $z$ and $\bar{z}$ and add.

Technique 6: Recognize regular polygons. Points $z_0, z_0\omega, z_0\omega^2, \ldots, z_0\omega^{n-1}$ (for $\omega = e^{2\pi i/n}$) form a regular $n$-gon. Any time you see equally spaced angles, the roots of unity polynomial $z^n - c = 0$ is probably lurking.

Technique 7: Convert geometry into multiplication. Collinearity, perpendicularity, concyclicity, and angle bisectors all have clean complex-number characterizations. Three points $A, B, C$ are collinear iff $\frac{C-A}{B-A} \in \mathbb{R}$. They are concyclic iff $\frac{(C-A)(D-B)}{(C-B)(D-A)} \in \mathbb{R}$.

Technique 8: Use arguments to prove collinearity and cyclicity. $\arg\!\left(\frac{z_2 - z_1}{z_3 - z_1}\right) = 0$ or $\pi$ means $z_1, z_2, z_3$ are collinear. Four points are concyclic iff $\arg\!\left(\frac{z_3 - z_1}{z_3 - z_2}\right) \equiv \arg\!\left(\frac{z_4 - z_1}{z_4 - z_2}\right) \pmod{\pi}$, i.e., the inscribed angles subtended by chord $z_1 z_2$ from $z_3$ and $z_4$ agree modulo $\pi$. This is equivalent to Technique 7's cross-ratio condition.

9. Worked AIME-Style Problems

Problem 1. Compute $(1 - i\sqrt{3})^{10}$.

Solution. Note $1 - i\sqrt{3} = 2e^{-i\pi/3}$. Then $(1-i\sqrt{3})^{10} = 2^{10}e^{-10i\pi/3} = 1024\,e^{-4i\pi/3} = 1024\!\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) = -512 + 512\sqrt{3}i$.

Problem 2. Find all $z$ with $|z| = 1$ and $z^6 + z^4 + z^2 + 1 = 0$.

Solution. Factor: $z^6+z^4+z^2+1 = (z^2+1)(z^4+1)$. Roots of $z^2 = -1$: $z = \pm i$. Roots of $z^4 = -1 = e^{i\pi}$: $z = e^{i\pi/4}, e^{3i\pi/4}, e^{5i\pi/4}, e^{7i\pi/4}$. All six roots lie on the unit circle.

Problem 3. The vertices of a regular hexagon are the roots of $z^6 = 64$. Find the sum of the squares of the side lengths.

Solution. The vertices are $2e^{i\pi k/3}$ for $k = 0,\ldots,5$. Compute $|1-e^{i\pi/3}|^2 = (1-1/2)^2 + (\sqrt{3}/2)^2 = 1/4 + 3/4 = 1$. So each side has length $|2||1 - e^{i\pi/3}| = 2$, and the sum of squares is $6 \cdot 4 = \boxed{24}$.

Problem 4. Let $\omega = e^{2\pi i/7}$. Compute $|\omega + \omega^2 + \omega^4|^2$.

Solution. Let $S = \omega + \omega^2 + \omega^4$ and $\bar{S} = \omega^6 + \omega^5 + \omega^3$. Since $1+\omega+\omega^2+\cdots+\omega^6=0$, we have $S+\bar{S}=-1$. Now compute

Reducing exponents modulo $7$, this becomes

Therefore $|S|^2=S\bar{S}=\boxed{2}$.

Problem 5. Find $\sum_{k=1}^{12} \cos\!\left(\frac{2\pi k}{13}\right)\cos\!\left(\frac{4\pi k}{13}\right)$.

Solution. Use the product-to-sum identity to get $\frac{1}{2}\sum_{k=1}^{12}\cos\!\left(\frac{6\pi k}{13}\right) + \frac{1}{2}\sum_{k=1}^{12}\cos\!\left(\frac{2\pi k}{13}\right)$. Each sum equals $\text{Re}\!\left(\sum_{k=1}^{12}\omega^{3k}\right)$ or $\text{Re}\!\left(\sum_{k=1}^{12}\omega^k\right)$ for $\omega = e^{2\pi i/13}$. Since neither exponent is divisible by 13, both geometric sums equal $-1$. Answer: $-1$.

10. Proof Problems

Problem 1. Let $A$, $B$, $C$ be the vertices of an equilateral triangle in the complex plane. Prove that $A^2 + B^2 + C^2 = AB + BC + CA$.

Solution. The expression

$A^2+B^2+C^2-AB-BC-CA$

is unchanged if the same complex number is added to $A$, $B$, and $C$, because all translation terms cancel. Therefore we may translate the triangle so its centroid is at the origin, giving $A+B+C=0$. Expanding $(A+B+C)^2=0$ yields

$A^2+B^2+C^2+2(AB+BC+CA)=0.$

It remains to show $AB+BC+CA=0$.

An equilateral triangle is invariant under $120^\circ$ rotation about its centroid. With the centroid at the origin and after labeling the vertices in counterclockwise order, we may write $B=\omega A$ and $C=\omega^2A$ where $\omega = e^{2\pi i/3}$. Then: $AB + BC + CA = \omega A^2 + \omega^3 A^2 + \omega^2 A^2 = A^2(1 + \omega + \omega^2) = 0,$ since $1 + \omega + \omega^2 = 0$ for any primitive cube root of unity. Hence $A^2 + B^2 + C^2 = -2(AB+BC+CA) = 0$ as well. $\square$

Problem 2. Show that if $|z_1| = |z_2| = |z_3| = 1$ and $z_1 + z_2 + z_3 = 0$, then $z_1, z_2, z_3$ form an equilateral triangle.

Solution. Since $|z_k|=1$, the squared distance is $|z_j - z_k|^2 = 2 - (z_j\bar{z}_k + \bar{z}_jz_k)$. Set $a_{12} = z_1\bar{z}_2 + \bar{z}_1z_2$, $a_{13} = z_1\bar{z}_3 + \bar{z}_1z_3$, $a_{23} = z_2\bar{z}_3 + \bar{z}_2z_3$.

Multiply $z_1+z_2+z_3 = 0$ by $\bar{z}_1$ to get $1 + z_2\bar{z}_1 + z_3\bar{z}_1 = 0$; adding its conjugate gives $2 + a_{12} + a_{13} = 0$. The analogous equations from multiplying by $\bar{z}_2$ and $\bar{z}_3$ yield $2 + a_{12} + a_{23} = 0$ and $2 + a_{13} + a_{23} = 0$. Solving this linear system gives $a_{12} = a_{13} = a_{23} = -1$. Therefore $|z_j - z_k|^2 = 3$ for all pairs, and the triangle is equilateral. $\square$

Problem 3. Distinct points $P, Q, R$ lie on the unit circle. Prove that $P, Q, R$ are the vertices of an equilateral triangle if and only if $P^3 = Q^3 = R^3$.

Solution. $(\Rightarrow)$ The circumcenter of any triangle inscribed in the unit circle is the origin. For an equilateral triangle, the circumcenter coincides with the centroid, so the centroid is at the origin. The triangle is therefore invariant under $120^\circ$ rotation about the origin, meaning $\{P,Q,R\} = \{P, e^{2\pi i/3}P, e^{4\pi i/3}P\}$. In either orientation, $Q^3 = e^{\pm 2\pi i}P^3 = P^3$ and $R^3 = e^{\pm 4\pi i}P^3 = P^3$. $(\Leftarrow)$ If $P^3 = Q^3 = R^3 = c$ with $|c|=1$, then $P$, $Q$, $R$ are three distinct cube roots of $c$. The cube roots of any $c$ on the unit circle are equally spaced at $120^\circ$, forming an equilateral triangle. $\square$

11. Common Mistakes

Forgetting multiple arguments. The argument of $z$ is $\theta + 2\pi k$, not just $\theta$. When taking $n$th roots, each value of $k$ from $0$ to $n-1$ gives a distinct root. Missing any of them loses points.

Losing roots. $z^n = w$ has exactly $n$ roots. If you find fewer, you dropped some. Always list all $n$.

Misapplying principal argument. $\text{Arg}(z_1 z_2) = \text{Arg}(z_1) + \text{Arg}(z_2)$ only holds modulo $2\pi$; the principal argument of a product may differ from the sum of principal arguments by $\pm 2\pi$.

Confusing modulus and argument. $|z_1 z_2| = |z_1||z_2|$ (multiply), but $\arg(z_1 z_2) = \arg(z_1) + \arg(z_2)$ (add). Swapping these is a fatal error.

Incorrect angle arithmetic. When computing $e^{i\theta} \cdot e^{i\phi} = e^{i(\theta+\phi)}$, be careful about angle reduction modulo $2\pi$. The form $re^{i\theta}$ is not unique unless you fix a range for $\theta$.

Forgetting conjugates on the unit circle. If $|z| = 1$, then $\bar{z} = 1/z$, not just $\overline{a+bi} = a - bi$. Using this saves enormous computation.

12. Practice Problems

Easy

1. Compute $(1 + i\sqrt{3})^6$.
2. Find all complex numbers $z$ with $z^4 = -16$.
3. If $\omega = e^{i\pi/5}$, simplify $\omega^3 \cdot \omega^7$.

Medium

4. Let $z = \frac{1+i}{1-i}$. Compute $z^{100}$.
5. Find $\sum_{k=0}^{5}\left(\cos\frac{\pi k}{3} + i\sin\frac{\pi k}{3}\right)$.
6. The complex number $z$ satisfies $|z|=2$ and $\arg(z) = \pi/4$. Find all $w$ with $w^3 = z$.
7. Compute $|\,(2+2i)^{10} - (2-2i)^{10}\,|$.

Hard

8. Let $\omega = e^{2\pi i/9}$. Compute $(\omega + \omega^8)(\omega^2 + \omega^7)(\omega^3+\omega^6)(\omega^4+\omega^5)$.
9. Prove that the product of all distances from $1$ to the other $n$th roots of unity equals $n$.
10. Find all $z$ on the unit circle such that $z^n + \bar{z}^n$ is an integer for every positive integer $n$.

13. Hints

1. Convert to polar form: $1 + i\sqrt{3} = 2e^{i\pi/3}$.
2. Write $-16 = 16e^{i\pi}$ and apply the root formula.
3. Add exponents and reduce modulo $2\pi$.
4. First simplify $\frac{1+i}{1-i}$ by multiplying numerator and denominator by $1+i$.
5. Recognize this as a sum of 6th roots of unity.
6. Take the cube root of the modulus; divide the argument by $3$; add $2\pi k/3$ for $k=0,1,2$.
7. Write both in polar form; note $2+2i = 2\sqrt{2}\,e^{i\pi/4}$.
8. Consider the polynomial with roots $\omega^k + \omega^{9-k}$ and use Vieta's or the minimal polynomial of $2\cos(2\pi/9)$.
9. Factor $z^n - 1 = \prod_{k=0}^{n-1}(z - \omega^k)$ and set $z = 1$ after canceling the $k=0$ factor.
10. Write $z = e^{i\theta}$; then $z^n + \bar{z}^n = 2\cos(n\theta)$. Consider when $2\cos(n\theta) \in \mathbb{Z}$ for all $n$.

14. Solutions

1. $(1+i\sqrt{3})^6 = (2e^{i\pi/3})^6 = 64e^{2\pi i} = 64$.

2. $-16 = 16e^{i\pi}$. Roots: $2e^{i(\pi + 2\pi k)/4}$ for $k=0,1,2,3$. These are $\sqrt{2}(1+i), \sqrt{2}(-1+i), \sqrt{2}(-1-i), \sqrt{2}(1-i)$.

3. $\omega^{3+7} = \omega^{10} = e^{i\cdot 10\pi/5} = e^{2\pi i} = 1$.

4. $\frac{1+i}{1-i} = \frac{(1+i)^2}{2} = i$. Then $i^{100} = (i^4)^{25} = 1$.

5. The terms are $e^{i\pi k/3}$ for $k=0,\ldots,5$, i.e., all 6th roots of unity. Their sum is $0$.

6. $z = 2e^{i\pi/4}$, so $w = 2^{1/3}e^{i(\pi/12 + 2\pi k/3)}$ for $k=0,1,2$.

7. $(2+2i)^{10} = (2\sqrt{2})^{10}e^{10i\pi/4} = 2^{15}e^{5i\pi/2} = 2^{15}i$. Similarly $(2-2i)^{10} = -2^{15}i$. Difference: $2^{16}i$, absolute value $\mathbf{65536}$.

8. Since $\omega^9 = 1$, each factor satisfies $\omega^k + \omega^{9-k} = 2\cos(2\pi k/9)$. Note the third factor: $\omega^3 + \omega^6 = 2\cos(2\pi/3) = -1$. So the full product equals $2\cos(2\pi/9) \cdot 2\cos(4\pi/9) \cdot (-1) \cdot 2\cos(8\pi/9) = -8\cos(2\pi/9)\cos(4\pi/9)\cos(8\pi/9)$. The three values $2\cos(2\pi/9), 2\cos(4\pi/9), 2\cos(8\pi/9)$ are roots of $x^3 - 3x + 1 = 0$; by Vieta's formulas their product is $-1$, so $\cos(2\pi/9)\cos(4\pi/9)\cos(8\pi/9) = -1/8$. The answer is $-8 \cdot (-1/8) = \mathbf{1}$.

9. From $z^n - 1 = \prod_{k=0}^{n-1}(z-\omega^k)$, cancel the $k=0$ factor to get $\prod_{k=1}^{n-1}(z-\omega^k) = z^{n-1}+\cdots+1$. Set $z=1$: $\prod_{k=1}^{n-1}(1-\omega^k) = n$. Taking absolute values gives the desired product of distances.

10. Write $z=e^{i\theta}$. The condition becomes

$z^n+\bar z^n=e^{in\theta}+e^{-in\theta}=2\cos(n\theta)\in\mathbb{Z}$

for every positive integer $n$. Taking $n=1$ already forces $2\cos\theta\in\mathbb{Z}$. Since $\cos\theta\in[-1,1]$, this gives

$2\cos\theta\in\{-2,-1,0,1,2\}.$

Thus

$\theta\in\{0,\pm\pi/3,\pm\pi/2,\pm2\pi/3,\pi\}.$

These values are also sufficient. If $x=2\cos\theta$ is an integer, then the recurrence

$2\cos(n\theta)=x\cdot 2\cos((n-1)\theta)-2\cos((n-2)\theta)$

shows by induction that $2\cos(n\theta)$ is an integer for all positive integers $n$. Therefore

$z\in\{1,e^{\pm i\pi/3},\pm i,e^{\pm 2i\pi/3},-1\}.$

These are exactly the sixth roots of unity together with $\pm i$.

15. Summary

Polar Form and De Moivre Decision Checklist

Use this checklist when encountering any complex number problem in a contest. The key question is always: does the structure of the problem favor angle addition (polar) or direct manipulation (rectangular)?

Use polar form when: the problem involves products, quotients, or powers of complex numbers; a geometric transformation is described; angles or rotations appear explicitly; or all given quantities lie on a circle.

Use De Moivre when: a high power of a complex number is requested; a trigonometric identity for $\cos(n\theta)$ or $\sin(n\theta)$ is needed; or a sum of the form $\sum \cos(k\theta)$ must be computed.

Use rotations when: the problem mentions equilateral triangles, squares, or regular polygons; a geometric locus is described as an orbit under rotation; or the phrase "rotate by angle $\theta$" appears.

Switch away from rectangular when: you see $a^2 + b^2$ (this is $|z|^2$), or $\cos\theta + i\sin\theta$ (this is $e^{i\theta}$), or you are multiplying two complex numbers and care about the angle of the result.

Common contest patterns:

• Equilateral triangle: $C - A = e^{\pm i\pi/3}(B-A)$
• Square: $C - B = i(B-A)$
• Regular $n$-gon: vertices are $c + re^{2\pi ik/n}$
• Roots of unity filter: $\sum_{k=0}^{n-1}\omega^{jk} = [n \mid j] \cdot n$
• Sum of all $n$th roots: always zero for $n \geq 2$
• Product of all distances from one root to the others: equals $n$