Draw specified set on a complex plane

Question

Draw specified set on a complex plane:

$\left| z+\left( \frac{1-i \sqrt{3} }{ \sqrt{2} + i \sqrt{2} } \right) ^{36} \right| \ge \left| z+i\right|$

I am not sure how to do this, tried to calculate trigonometric form of expression in bracket, but failed. Sorry for any spelling mistakes I made, English is not my primary language.

Answer 1

in a first step prove that $$\left(\frac{1-i\sqrt{3}}{\sqrt{2}+i\sqrt{2}}\right)^{30}=i$$

Answer 2

Hint:

This is the same as $$\biggl|z+\Bigl(\mathrm e^{-\tfrac{i\pi}{12}}\Bigr)^{36}\biggr|=\biggl|z-1\biggr|\ge |z+i|. $$ Interpret in terms of distance of $z$ to $1$ and $i$ respectively: this describes the half-plane above the perpendicular bisector of the segment joining the images of $i=e^{\tfrac{i\pi}{2}}$ and $1$; i.e. the line with (complex) equation: $\;\arg z=\frac{\pi}{4}$.

Answer 3

Hint $\frac{1-i\sqrt 3}{\sqrt 2 +i\sqrt 2}=\frac{e^{i5\pi/6}}{e^{i\pi/4}}$

Answer 4

Hint: $\frac{1-i\sqrt{3}}{\sqrt{2}+i\sqrt{2}} = \frac{e^{-i\pi/3}}{e^{i\pi/4}} = e^{-i7\pi/12}$ elevate that to the power 36 and you get $|z+e^{-i\pi}|\geq|z+i|$

So $|z-1|\geq|z+i|$

Interpret in terms of distance between $z$ to $1$ and $-i$ respectively.