The problem given read as follows:

Using complex numbers, find all the vertices of a regular polygon of $n$ sides knowing that its center is located at $z = 0$ and one of its vertices lies in $z_{1} = i$.

I would like to know if my answer is correct. Here's what I have thought:

We have a regular polygon centred at $(0,0)$ and a known affix $z_{1} = 0 + i = (0, 1)$ in the complex plane. The n-th roots of unity will compose a circle with the same center as the polygon such that all the vertices generate a regular polygon inscribed in the circle $|z| = 1$.

In polar form, we can write $z_{1} = i = e^{i \frac{\pi}{2}}$. The next vertex, in turn, will be located at an angle with $\frac{2 \pi}{n}$ of $z_{1}$ and so on and so forth. From DeMoivre's formula, we know that the roots of unity are given by:

$z_{k} = |z|^\frac{1}{n}.cis \left(\theta + 2k\pi\right), k \in \mathbb{R}$, with $ 0 \leq k \leq n - 1$. In this particular case, $\theta = \frac{\pi}{2}$.

So the first vertex would be $z_{1}=cis \left(\frac{\pi}{2} \right)$, the second vertex $z_{2} = cis \left(\frac{\pi}{2} + \frac{2 \pi}{n} \right)$, till $z_{n} = cis \left(\frac{\pi}{2} + \frac{2(n-1) \pi}{n} \right)$.

It is just that?

Thanks, in advance.

  • $\begingroup$ You are not keeping track of your n's and your k's. $z_{k} = cis \left(\frac{\pi}{2} + \frac {2k\pi}{n} \right)$ $\endgroup$
    – Doug M
    Commented Sep 8, 2017 at 21:29
  • $\begingroup$ Thanks, I had forgotten the n in the denominator while typing. $\endgroup$
    – MCarsten
    Commented Sep 8, 2017 at 21:32


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