Find all complex numbers $z$ such that ($z^6 - i) \in \mathbb R$ 
Find all complex numbers $z$ such that ($z^6 - i) \in \mathbb R$

My solution: 

Let's set $x^6 = (z - i)^6$. Then
$$x^6 = |x| e^{6\theta i} \\
x^6 \in \mathbb R \iff 6\theta = k\pi \land k\in \mathbb Z$$
$$\theta = \frac{k\pi}{6}$$
Therefore $z - i = |z - i|(\cos(\frac{k\pi}{6}) + i\sin(\frac{k\pi}{6})) \\$
$$z = |z-i|\left(\cos\left(\frac{k\pi}{6}\right)+i\sin\left(\frac{k\pi}{6}\right)\right)+i$$
Now, imagine that I have plotted the solution in terms of $x$. If I wanted to have a plot in terms of $z$, would it be enough to simply shift all of my solutions one imaginary unit upwards, to satisfy the $+i$ term? 


$z^6 = \cot \theta + i\\
z^6 = \csc \theta (\cos \theta + i\sin\theta)\\
z = (\csc \theta)^{\frac 16} e^{i(\frac {\theta}{6}+\frac {k\pi}{3})}\\$
