Maximum value of $|z-\sqrt3-i|^2$

Question

If $S=\{z\in C:\overline z=iz^2\}$, then the maximum value of $|z-\sqrt3-i|^2$ on $S$ is:

I took $z$ as $x+iy$ then $S$ came out to be $\{(0,0),(0,1),(\frac{\sqrt3}{2},\frac{-1}{2}), (\frac{-\sqrt3}{2},\frac{-1}{2})\}$. Taking their distances from $(\sqrt3,-1)$, I get maximum value as $7$ but the answer is given as $9$.

Answer 1

I think you made a mistake here:the distance you need to find is from $(\sqrt{3},1)$ and the distance is maximum for the point ($-\frac{\sqrt{3}}{2},-\frac{1}{2})$) which is 3 and aquare of that is 9

Answer 2

Yes, the answer is $9$. You have$$S=\left\{0,i,\frac{\sqrt3}2-\frac12i,-\frac{\sqrt3}2-\frac12i\right\}$$and the square of the distance from each element of $S$ to $\sqrt3+i$ is equal to $4$, $3$, $3$, and $9$ respectively.

Answer 3

Since you are also asking for a quicker way in one of your comments.

You may first derive that besides $z=0$ as obvious solution, all other solutions satisfy $|z|=1$:

$$z\neq 0:\; \bar z = i z^2 \Rightarrow |\bar z|= |z|=|z|^2\Rightarrow |z|=1$$

Now, solve $$e^{-it} = e^{(2t+\frac{\pi}{2})i}\Leftrightarrow 3t+\frac{\pi}2=2k\pi$$

So, you get as solutions (modulo $2\pi$)

$$t=-\frac{\pi}{6},\; t=\frac{\pi}{2},\; t=\frac{7\pi}{6}$$

Rewriting $e^{it}$ in coordinate form gives the non-zero entries in your set $S$.