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:
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Find the solutions to $z^n - 1 = 0,~ z \in \mathbb{C}$.
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Plot these solutions on the Argand plane.
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Draw a tree diagram starting from the trivial solution $z = 1$ to every other root.
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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 –
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The solutions to $z^n - 1 = 0$ are obtained most easily through Euler’s form: $$ z_k = \exp\left(\frac{2k\pi i}{n}\right),~~k = 0,…,n-1 $$
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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:
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The easiest observation is that the sum of the roots is zero: $$ \sum^{n-1}_{k = 0} z_k = 0 $$
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The roots of unity can be factorised into the following irreducible form over the field of real numbers: $$ z^n - 1 = (z-1)\sum^{n-1}_{i = 0} z^i $$ and can be factorised over the complex field as: $$ z^n - 1 = \prod_{k=0}^{n-1}(z-z_k) = (z-1)\prod_{k=1}^{n-1}(z-z_k) $$
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A conjecture based on trial and error is: $$ \prod_{k=1}^{n-1}|z_k - z_0| = \prod_{k=1}^{n-1}|1-z_k| = n$$ 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: $$\sum^{n-1}_{i = 0} z^i = \prod_{k=1}^{n-1}(z-z_k) = \frac{z^n - 1}{z-1}$$ For the equalities to be sensible at $z=1$, one must perform analytic continuation: $$\sum^{n-1}_{i = 0} 1^i = \prod_{k=1}^{n-1}|1-z_k| = \lim_{z\rightarrow 1} \frac{z^n - 1}{z-1}$$ The first equality proves the conjecture. The last equality is evaluated using L’Hôpital’s rule, also proving the conjecture: $$ \lim_{z\rightarrow 1} \frac{z^n - 1}{z-1} = \lim_{z\rightarrow 1} nz^{n-1} = n $$