Finding conjugates of all z $\in$ C that satisfy $z^3$

Question

$$ z^3 = \frac{16e^{i\frac{3\pi}4}}{(1-\sqrt3)+\sqrt6e^{i\frac\pi4}} $$

Anyone know of a good way to simplify this expression?

Answer 1

$\sqrt{6}e^{i\pi/4} = \sqrt{6}\left(\cos(\pi/4) + i\sin(\pi/4)\right) = \sqrt{6}\left(\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2}i\right) = \sqrt{3} + \sqrt{3}i \Rightarrow 1 - \sqrt{3} + \sqrt{6}e^{i\pi/4} = 1 +\sqrt{3}i = 2e^{i\pi/3} \Rightarrow z^3 = \dfrac{16e^{i3\pi/4}}{2e^{i\pi/3}} = 8e^{i5\pi/12} = 8e^{i5\pi/12+2n\pi} \Rightarrow z = 2e^{i5\pi/36 + 2n\pi/3} \Rightarrow \bar{z} = 2e^{-i5\pi/36 - 2n\pi/3}, n \in \{0,1,2\}$.

Answer 2

$$e^{\frac{3\pi i}4}=-\frac1{\sqrt2}+\frac1{\sqrt2}i=\frac1{\sqrt2}(-1+i)$$

$$\sqrt6e^{\frac{\pi i}4}=\sqrt6\left(\frac1{\sqrt2}+\frac1{\sqrt2}i\right)=\sqrt3(1+i)$$

and thus your expression is

$$\frac{16\frac1{\sqrt2}(-1+i)}{1-\sqrt3+\sqrt3+\sqrt3 i}=-8\sqrt2\frac{1-i}{1+\sqrt3i}$$

Try to take it now from here.

Answer 3

$ (1-\sqrt3)+\sqrt6e^{i\frac\pi4}$ = $(1-\sqrt3)$+ $\sqrt6$($\frac{1}{\sqrt2}$ +$i\frac1{\sqrt2}$)$ = $2$e^{i\frac{\pi}{3}}$

$ z^3= 8e^{i\frac{5\pi}{12}}$