Why the Three Cube Roots of 1 Always Add Up to Zero

A real number sits on a line, but a complex number needs two directions — one real, one imaginary. Plotting it on a plane turns algebra into geometry: moduli become distances, arguments become angles, and multiplication becomes rotation.

This part covers the Argand plane and polar form, geometric interpretation, the cube roots of unity, and using their properties to simplify expressions.

**The complex number $z=a+ib$ is plotted as the point $(a,b)$ on the Argand plane, and writing $r=|z|$ and $\theta =\arg z$ gives the polar form $z=r(\cos \theta +i\sin \theta )$, with $a=r\cos \theta$ and $b=r\sin \theta$.

The Argand plane:

- The horizontal axis is the real axis and the vertical axis is the imaginary axis**
- $|z|$ is the distance of the point from the origin
- $\bar{z}$ is the reflection of z in the real axis

Worked example. Write $z=1-i\sqrt{3}$ in polar form.

- $r=\sqrt{1+3}=2$
- $\cos \theta =\frac{1}{2}$ and $\sin \theta =-\frac{\sqrt{3}}{2}$, so the point is in the fourth quadrant and $\theta =-\frac{\pi }{3}$

$$z=2\left(\cos \left(-\frac{\pi }{3}\right)+i\sin \left(-\frac{\pi }{3}\right)\right)$$

Worked example 2. Convert back: $4\left(\cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6}\right)=4\left(-\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)=-2\sqrt{3}+2i\approx -3.46+2i$.

An everyday example. Giving directions as 'walk 2 km at this angle' instead of 'go 1 km east and 1.73 km south' is exactly the switch from algebraic to polar form.

The substance. Polar form makes multiplication quick — moduli multiply and arguments add, so $z^2$ above has modulus 4 and argument $-\frac{2\pi }{3}$.

**On the Argand plane, $|z_1-z_2|$ is the distance between two points, addition follows the parallelogram law, multiplying by i rotates a point through $90°$ anticlockwise about the origin, and equations such as $|z-z_0|=r$ describe circles.

Key interpretations:**

- $z_1+z_2$ — the diagonal of the parallelogram formed by $z_1$ and $z_2$
- $|z_1-z_2|$ — the distance between the points $z_1$ and $z_2$
- $|z-z_0|=r$ — a circle with centre $z_0$ and radius r
- $|z-z_1|=|z-z_2|$ — the perpendicular bisector of the segment joining $z_1$ and $z_2$
- $iz$ — z rotated through $90°$ anticlockwise, so $i(3+i)=-1+3i$

Worked example. Find the distance between $z_1=2+3i$ and $z_2=-1-i$:

$$|z_1-z_2|=|3+4i|=\sqrt{9+16}=5$$

Worked example 2. The locus $|z-(1+2i)|=3$ is the circle $(x-1{)}^2+(y-2{)}^2=9$. The point $4+2i$ lies on it, since $|(4+2i)-(1+2i)|=|3|=3$.

An everyday example. A mobile tower serving every phone within 5 km covers the disc $|z-z_0|\le 5$ on a map drawn as an Argand plane.

The substance. Multiplying by i is a rotation — doing it four times brings a point back to where it started, matching $i^4=1$.

**The cube roots of unity are the solutions of $z^3=1$, namely $1$, $\omega =\frac{-1+i\sqrt{3}}{2}$ and ${\omega }^2=\frac{-1-i\sqrt{3}}{2}$, and they satisfy $1+\omega +{\omega }^2=0$ and ${\omega }^3=1$.

Finding them.**

$$z^3-1=(z-1)(z^2+z+1)=0$$

so $z=1$ or $z^2+z+1=0$, which gives $z=\frac{-1\pm i\sqrt{3}}{2}$.

Key properties:

- ${\omega }^3=1$, so powers of $\omega$ repeat every three: ${\omega }^{3k+r}={\omega }^r$
- Sum $1+\omega +{\omega }^2=0$, because $\omega$ is a root of $z^2+z+1=0$
- Product $1\times \omega \times {\omega }^2={\omega }^3=1$
- $\bar{\omega }={\omega }^2$ and $|\omega |=1$
- On the Argand plane they form an equilateral triangle on the unit circle, at arguments $0$, $\frac{2\pi }{3}$ and $-\frac{2\pi }{3}$

Numerical check. $\omega +{\omega }^2=\frac{(-1+i\sqrt{3})+(-1-i\sqrt{3})}{2}=-1$, so $1+\omega +{\omega }^2=0$.

An everyday example. **The three blades of a ceiling fan, spaced $120°$ apart, balance one another — just as the three cube roots of unity, pointing equally in three directions, add to zero.

The substance. ${\omega }^2$ is not $|\omega {|}^2$** — $|\omega {|}^2=1$, but ${\omega }^2$ is the third, non-real cube root.

**Expressions in $\omega$ are simplified by reducing powers with ${\omega }^3=1$ and replacing $1+\omega$ with $-{\omega }^2$, $1+{\omega }^2$ with $-\omega$, and $\omega +{\omega }^2$ with $-1$.

Useful substitutions:**

- $1+\omega =-{\omega }^2$
- $1+{\omega }^2=-\omega$
- $\omega +{\omega }^2=-1$

Worked example. Evaluate $(1-\omega +{\omega }^2)(1+\omega -{\omega }^2)$.

- $1+{\omega }^2=-\omega$, so $1-\omega +{\omega }^2=-2\omega$
- $1+\omega =-{\omega }^2$, so $1+\omega -{\omega }^2=-2{\omega }^2$
- Product $=(-2\omega )(-2{\omega }^2)=4{\omega }^3=4$

Worked example 2. Evaluate $(1+\omega -{\omega }^2{)}^6$. It equals $(-2{\omega }^2{)}^6=64{\omega }^{12}=64$.

Worked example 3. $x^2+x+1=(x-\omega )(x-{\omega }^2)$, which is why $x^2+x+1$ divides $x^3-1$.

An everyday example. Three-phase electricity supplied to Indian factories uses three voltages $120°$ apart that add to zero at every instant — the cube roots of unity at work.

The substance. Always reduce the exponent using division by 3 first — ${\omega }^{100}$ equals $\omega$, since $100=3\times 33+1$.