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? 
 A: 
Known, $w\in \Bbb R$ iff $w=\bar{w}$ 

Then set $z= re^{it}$
$$(z^6 - i)\in\Bbb R\Longleftrightarrow  (z^6 - i)= (\bar{z}^6 + i)\\\Longleftrightarrow (z^6 -\bar{z}^6 = 2 i)  \Longleftrightarrow Im(z^6) = 1\\ \Longleftrightarrow \color{blue}{r^6\sin (6t)  }=Im(r^6 e^{i6t}) = 1 $$

Conclusion $$(z^6 - i)\in\Bbb R \Longleftrightarrow \color{blue}{r^6\sin (6t) = 1}$$ with $z= re^{it}$

A: I am going to us Doug M.'s nice diagram.  Let $ a \ge 0$.  Then we are looking for 
$$
z^6=a+i
$$
Writing $a+i$ in polar form, we have
$$
z^6=\sqrt{a^2+1}e^{\theta_a i}
$$
where $\theta_a=arctan(1/a)$. Hence
$$
z=(a^2+1)^{1/12}e^{\theta_a i/6}
$$
We should also consider all the primitive roots of unity. Thus
$$
z=(a^2+1)^{1/12}e^{\theta_a i/6}e^{k\pi i /3}
$$
where $k \in \{0, 1,..., 5\}$
In particular, when $a=0$
$$
z=e^{\pi i/12}e^{k\pi i /3}
$$
We also have
$$
z^6 = -a+i
$$
in which case
$$
z=(a^2+1)^{1/12}e^{-\theta_a i/6}e^{k\pi i /3}
$$
where I have used the fact that $arctan(1/a)=-arctan(-1/a)$.
So there is a whole continuum of such $z$, parameterized by $a \in \mathbb{R}$ and $k \in \{0, 1, ..., 5\}$
A: 
$z^6 = \cot \theta + i\\
z^6 = \csc \theta (\cos \theta + i\sin\theta)\\
z = (\csc \theta)^{\frac 16} e^{i(\frac {\theta}{6}+\frac {k\pi}{3})}\\$
