if $z=2021+2021i$ prove $w_0+\cdots+ w_{2020}=0$

I'm trying to solve the following problem:

lets denote $w_0,\ldots, w_{2020}$ to be all the 2021 different roots of order $2021$ of the complex number $2021+2021i$. Prove that $w_0+\cdots+ w_{2020}=0$?

As a starter, I turned $z=2021+2021i$ to trigonometric form so I got $r=2021\sqrt{2}$ and $\theta=\frac{\pi}{4}$. This means that: $$ w_0+\cdots + w_{2021}=\sum_{k=0}^{2020}\operatorname{cis}\left(\frac{\frac{\pi}{4}+2\pi k}{2021}\right) $$ Now I'm stuck. I could turn it to be of form $e^{i\theta}$ but I don't see how it can help here. How can I evaluate it?

Solution 1:

The simplest solution (by far) is the very excellent one offered by Arthur Vause in the comments: if $z$ is a $2021^\text{st}$ root of $202i + 2021i$, then $z^{2021} - (2021 + 2020i) = 0$, and the coefficient of $z^{2020}$ in that polynomial is the sum of all the roots. (And since there is no $z^{2020}$ in the polynomial, its coefficient is ....)

But there are some other ways to approach the problem that may help you understand it better. First, your equation needs a correction, as you dropped the $r$ factor: $$\begin{align}w_0+\cdots + w_{2021}&=2021\sqrt2\sum_{k=0}^{2020}\operatorname{cis}\left(\frac{\frac{\pi}{4}+2\pi k}{2021}\right)\\ &=2021\sqrt2\sum_{k=0}^{2020}e^{i(\pi/4 + 2\pi k)/2021}\\ &=2021\sqrt2e^{i\pi/8084}\sum_{k=0}^{2020}\left(e^{2\pi i/2021}\right)^k\\& =w_0\sum_{k=0}^{2020}\omega^k\end{align}$$ where $\omega = e^{2\pi i/2021}$.

If you multiply $\sum_{k=0}^{2020}\omega^k$ by $\omega - 1$ (this is a very well-known calculation), the result is $$\sum_{k=0}^{2020}\omega^{k+1} - \sum_{k=0}^{2020}\omega^k = \omega^{2021} - \omega^0 = \omega^{2021} - 1$$ And therefore $$\sum_{k=0}^{2020}\omega^k = \dfrac{\omega^{2021} - 1}{\omega-1}$$ But $\omega^{2021} = \left(e^{2\pi i/2021}\right)^{2021} = e^{2\pi i} = 1$. So $$\sum_{k=0}^{2020}\omega^k = \frac{1 - 1}{\omega -1} = 0$$

A more geometric approach is to note that the $2021$ roots of unity, $\omega^k, 0\le k \le 2020$ are points equally spaced around the unit circle. Their average value should be the point that they are equally distributed about: the origin. But the average is just the sum divided by $2021$, so their sum has to be $0$ as well.

Of course, that is intuition rather than an actual proof. But if $2021$ were an even number, the fact that the sum is zero would be obvious, because every number in the sum would have its opposite in the sum as well. For odd numbers like $2021$, it is a little harder. But if you take a square root of $\omega$ (either square root will do) and add $$\sum_{k=0}^{2020}\omega^k + \sqrt \omega\sum_{k=0}^{2020}\omega^k = \sum_{k=0}^{4041}(\sqrt \omega)^k = 0$$ because it is the sum of all the $4042$ roots of unity. Since $1 + \sqrt \omega \ne 0$, this means $\sum_{k=0}^{2020}\omega^k$ must be $0$.