Power of complex number using exponential form $x=2(\cos(\pi/3) + i\sin(\pi/3), y=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} $
Question:
a.  Is there exists $k\in\mathbb{N}$ such that $Re((xy)^k) = 0$
My approach is,
a. $x=2e^{i\frac{\pi}3}, y=1(\cos(-\pi/4) + i\sin(-\pi/4))=1e^{-i\frac{\pi}{4}} $
Hence, $$(xy)^k = \left(2e^{-i\pi/4+\pi/3i}\right)^k=2e^{i\frac{\pi}{12}k}  = 2(\cos(\pi/12k) + i\sin(\pi/12k) $$, So
$$\frac{\pi}{12}k = \frac{\pi}{2} + l\pi$$
$$\frac{1}{12}k=\frac12 + l$$
$$k = 6 + 12l$$,
yes, there exists such $k$, for example $k=18$.  
Is it ok ?
 A: Generalize the problem, when $\text{z}_1\space\wedge\space\text{z}_2\in\mathbb{C}$, does a number $k$ exist for which $\Re\left[\left(\text{z}_1\text{z}_2\right)^k\right]=0$ where $k\in\mathbb{N}$:


*

*$$\text{z}_1=\Re[\text{z}_1]+\Im[\text{z}_1]i=|\text{z}_1|e^{\left(\arg(\text{z}_1)+2\pi n\right)i}$$
Where $|\text{z}_1|=\sqrt{\Re^2[\text{z}_1]+\Im^2[\text{z}_1]}$, $\arg(\text{z}_1)$ is the complex argument of $\text{z}_1$ and $n\in\mathbb{Z}$.

*$$\text{z}_2=\Re[\text{z}_2]+\Im[\text{z}_2]i=|\text{z}_2|e^{\left(\arg(\text{z}_2)+2\pi n\right)i}$$

*$$\text{Q}=\text{z}_1\text{z}_2=|\text{z}_1||\text{z}_2|e^{\left(\arg(\text{z}_1)+\arg(\text{z}_2)+2\pi n\right)i}$$
So:
$$\text{Q}^k=\left(|\text{z}_1||\text{z}_2|e^{\left(\arg(\text{z}_1)+\arg(\text{z}_2)+2\pi n\right)i}\right)^k=|\text{z}_1|^k|\text{z}_2|^ke^{k\left(\arg(\text{z}_1)+\arg(\text{z}_2)+2\pi n\right)i}$$
For the real part (using Euler's formula):
$$\Re\left[\text{Q}^k\right]=|\text{z}_1|^k|\text{z}_2|^k\cos\left(k\left(\arg(\text{z}_1)+\arg(\text{z}_2)+2\pi n\right)\right)=0$$
So, for $k$ we get the following solutions:


*

*$$|\text{z}_1|^k=0$$

*$$|\text{z}_2|^k=0$$

*$$\cos\left(k\left(\arg(\text{z}_1)+\arg(\text{z}_2)+2\pi n\right)\right)=0$$

