Finding the value of $\frac{1}{z_1-z_2}+\frac{1}{z_2-z_3}+...+\frac{1}{z_7-z_1}$

Question

If $z_1,...,z_7$ are the roots to the equation $z^7=1+\sqrt{3}i$ in order of increasing argument, I want to find the value of $\frac{1}{z_1-z_2}+\frac{1}{z_2-z_3}+...+\frac{1}{z_7-z_1}$.

My attempt: We know that $z_1-z_2$, $z_2-z_3$,...,$z_7-z_1$ are the side lengths of a heptagon which sum to give $z_7$ and whose vertices lie on a circle of radius $2$.

If I consider the map of $z$ to $\frac{1}{z}$, whereby I divide by $|z|^2$ and then take the conjugate, I see that circles are preserved on the $\frac{1}{z}$ plane except the points are reflected about the real axis and the radius is reduced by $|z|^2$.

Graphically I have a feeling that $\frac{1}{z_1-z_2}+\frac{1}{z_2-z_3}+...+\frac{1}{z_7-z_1}=0$ by trying to picture the transformation but I'm not sure how to justify it using solid mathematical reasoning (if it is even the correct value).

So could someone show me how to find the correct value with my line of reasoning, if it is possible, or show me if there are any other more elegant and faster ways?

Answer 1

As you observed, the differences $z_1-z_2, z_2-z_3, \ldots, z_7-z_1$ are the side length of a regular heptagon, so they all have the same length, let's call that $L$. Then $$ \frac{1}{z_1-z_2}+\frac{1}{z_2-z_3}+ \cdots+\frac{1}{z_7-z_1} = \frac{\bar z_1- \bar z_2}{L^2} + \frac{\bar z_2- \bar z_3}{L^2} + \cdots + \frac{\bar z_7- \bar z_1}{L^2} \\ = \frac{1}{L^2}(\bar z_1- \bar z_2+\bar z_2- \bar z_3 + \cdots + \bar z_7- \bar z_1) = 0 \, . $$

Answer 2

$Z=2^{1/7}[\cos(\frac{2kπ+\frac{π}{3}}{7})+\sin(\frac{2kπ+\frac{π}{3}}{7})]$ where $k= 0,1,2,..., 6$

Putting the values of $k$ you can obtain the required result