During my second year of the International Baccalaureate: Diploma Programme, we were assigned the task of making a portfolio for a problem statement in Mathematics HL. While I don’t have the problem statement anymore, I’ll outline the major points of the task:

Find the solutions to $z^n  1 = 0,~ z \in \mathbb{C}$.

Plot these solutions on the Argand plane.

Draw a tree diagram starting from the trivial solution $z = 1$ to every other root.

Investigate the exercise, devise a conjecture and prove it.
When I first read through this exercise, I didn’t really expect anything interesting to show up. Well, I was horribly wrong; something really cool showed up which made me learn about a lot of little things in complex analysis as a result. I’ll start by solving the above points sequentially –

The solutions to $z^n  1 = 0$ are obtained most easily through Euler’s form: $$ z_k = \exp\left(\frac{2k\pi i}{n}\right),~~k = 0,…,n1 $$

As they are the roots of unity, they appear as points on the unit circle in the Argand plane. For example, $z^5  1 = 0$ has the following solutions and tree diagram:

The easiest observation is that the sum of the roots is zero: $$ \sum^{n1}_{k = 0} z_k = 0 $$

The roots of unity can be factorised into the following irreducible form over the field of real numbers: $$ z^n  1 = (z1)\sum^{n1}_{i = 0} z^i $$ and can be factorised over the complex field as: $$ z^n  1 = \prod_{k=0}^{n1}(zz_k) = (z1)\prod_{k=1}^{n1}(zz_k) $$

A conjecture based on trial and error is: $$ \prod_{k=1}^{n1}z_k  z_0 = \prod_{k=1}^{n1}1z_k = n$$ This states that the product of the distances between each root from a selected root is equal to the number of roots. To prove this, notice that: $$\sum^{n1}_{i = 0} z^i = \prod_{k=1}^{n1}(zz_k) = \frac{z^n  1}{z1}$$ For the equalities to be sensible at $z=1$, one must perform analytic continuation: $$\sum^{n1}_{i = 0} 1^i = \prod_{k=1}^{n1}1z_k = \lim_{z\rightarrow 1} \frac{z^n  1}{z1}$$ The first equality proves the conjecture. The last equality is evaluated using L’Hôpital’s rule, also proving the conjecture: $$ \lim_{z\rightarrow 1} \frac{z^n  1}{z1} = \lim_{z\rightarrow 1} nz^{n1} = n $$