Complex numbers on the Argand diagram

What you are looking at

In modulus–argument form, multiplying two complex numbers multiplies their moduli and adds their arguments: $|zw| = |z|\,|w|$ and $\arg(zw) = \arg z + \arg w$, adjusted by $2\pi$ where needed to stay in $(-\pi, \pi]$. The diagram draws $z$ in blue and $w$ in green, each with an arc marking its argument, and the product in yellow with its own arc. Multiplying by $w$ therefore scales by $|w|$ and rotates by $\arg w$, which you can watch happen as you drag.

The equation $z^n = 1$ has exactly $n$ solutions, $z = \operatorname{cis}\left(\dfrac{2\pi k}{n}\right)$ for $k = 0, 1, \dots, n-1$. They all lie on the unit circle, $\dfrac{2\pi}{n}$ apart, so they are the vertices of a regular $n$-gon with one vertex at $1$. Each root is a power of $\omega = \operatorname{cis}\left(\dfrac{2\pi}{n}\right)$, and the working panel shows that $1 + \omega + \omega^2 + \dots + \omega^{n-1} = 0$, which the symmetry of the polygon makes easy to believe.

Try this

1. 1Drag $w$ onto $i$, the point $(0, 1)$. Now $|w| = 1$ and $\arg w = \dfrac{\pi}{2}$, so $zw$ is exactly $z$ turned a quarter-turn anticlockwise about the origin.
2. 2Drag $w$ inside the dashed unit circle. With $|w| < 1$ the product lands closer to the origin than $z$ does.
3. 3Press Rotate. $w$ travels once round a circle at its own modulus, and $zw$ follows round a circle of radius $|z|\,|w|$.
4. 4Switch to Roots of unity and set $n = 6$. The roots form a regular hexagon with neighbours $\dfrac{\pi}{3}$ apart; press Step round to move the selected root one place at a time, or tap a root to select it.

Mistakes students make

• Finding an argument with $\arctan\dfrac{b}{a}$ alone. The starting product $-0.45 + 2.3i$ lies in the second quadrant, so its argument is $\pi - \arctan\dfrac{2.3}{0.45} \approx 1.76$, not $\arctan\left(\dfrac{2.3}{-0.45}\right) \approx -1.38$.
• Adding the moduli instead of multiplying them. With $|z| \approx 1.80$ and $|w| = 1.3$, the product has modulus about $2.34$, not $3.10$.
• Missing the root $z = 1$. The $n$th roots of unity start at $k = 0$, so there are $n$ of them, not $n - 1$, and neighbouring roots are $\dfrac{2\pi}{n}$ apart.

Where it is taught

• IB Maths AA HL 1.12 Complex numbers and the Argand diagram
• IB Maths AA HL 1.13 Modulus–argument form; products and their geometric interpretation
• IB Maths AA HL 1.14 Powers and roots of complex numbers