If $z_1,z_2\in \mathbb{C}$ are two consecutive points on a regular n-gon. Find the rest of the points.

Question

I did a question where I was given points $z_0, z_1$ where $z_0$ was the circum-radius and $z_1$ was a point on the regular n-gon. There, the rest of the points were found as $$z_k=z_0+(z_1-z_0)e^{\frac{2\pi k i}{n}},k=0,1,...,n-1$$ This is clear to me. (translation and rotation in $\mathbb{R^2}$ basically )

Then in the answer to the question I post, in the answer section it says that from this formula above that I state ,it can be deducted that: $$z_0=\frac{z_1+z_2}{2}\pm i\left(\frac{z_2-z_1}{2}\right)\cot\frac{\pi}{n}$$ I cannot see this myself. Can anyone help out with this?

Answer 1

If nothing else, the brute-force method must work. The first formula gives $z_2$ in terms of $z_1$ and $z_0$. Substitute that into the second formula. Write $w=e^{2 \pi i/n}$; write the cot in terms of exponentials; convert all exponential terms into powers of $w$. The right hand side is then a rational function in $z_0$, $z_1$, $w$. Expand everything out and simplify. If the formula is in fact true, this must work.

Answer 2

Substitute k=2 in your formula and divide by $z_1-z_0$ to get $\frac{z_2-z_0}{z_1-z_0} = e^{\frac{4\pi i}{n}}$. Now apply componendo dividendo rule of elementary algebra both sides and simplify to get $\frac{z_2+z_1-2z_0}{z_2-z_1} = \frac{1+e^{\frac{4\pi i}{n}}}{1-e^{\frac{4\pi i}{n}}}$. Now, simplify RHS.