Complex numbers-polygon

Question

I have tried finding the solution but i can not seem to find it. Thank you in advance for the answer!

Answer 1

This is fairly straightforward if you think in terms of the polar notation for complex numbers:

Every complex number $z$ can be expressed as $z = r \cdot e^{i \theta}$, where $r$ is its norm (the "radius") and $\theta$ is its argument (the angle between the real axis and the vector from the origin to $z$ in the complex plane). Note that if $\theta$ is that angle, any $\theta' = \theta + 2\pi k$ works as well, so feel free to follow any convention (e.g. $\theta \in [0, 2\pi)$).

Now, onto the question: a number $z_0 \in \mathbb{C}$ has in fact $n$ $n$-th roots $\sqrt[n]{z_0} = z_0^{\frac{1}{n}}$ in the complex numbers, which are precisely those $x = r_x \cdot e^{i\theta_x} \in \mathbb{C}$ which satisfy $x^n = z_0$. Writing $z_0 = r_z \cdot e^{i\theta_z}$, we have: $$x^n = r_x^n \cdot e^{ni\theta_x} = r_z \cdot e^{i\theta_z}.$$

In other words, our complex roots are the x satisfying $$r_x^n = r_z, \quad n\theta_x = \theta_z.$$

The first equation tells us that all of our solutions have the same norm $\sqrt[n]{r_z}$ and therefore lie on the circumference with centre at the origin and radius $\sqrt[n]{r_z}$.

The second equation is in fact an equation modulo $2\pi$, and its solutions are $\theta_x = \frac{\theta_z}{n} + k\frac{2\pi}{n}$. Note that a set of representatives of those angles is given by taking $k = 0, 1, ..., n-1$.

This gives us $n$ roots at angles $\frac{\theta_z}{n} + k\frac{2\pi}{n}$, and since those angles split the $2\pi$ radians of the circle evenly, the polygon formed by the roots is regular.

You can find a visual example here