Roots of $(z^2-z)^4=81$ in polar form Question :
Find the roots of $(z^2-z)^4=81$ in polar form.
What i've done :
I'm thinking about using De Moivre's Theorem.
Let $w=z^2-z \Rightarrow w^4=81$ 
$\begin{align}
&\Leftrightarrow w^4=3^4\,\text{cis $2\pi$}\\
&\Leftrightarrow w=3\,\text{cis $\left(\dfrac{2\pi+2k\pi}{4}\right)$}\\
&k=0,1,2,3
\end{align}$
What's next? I'm confusing to substitute $w=z^2-z$ back...
Please help me. Thanks
 A: I like your idea that first, you denote $$w=z^2-z.$$
Then, recall that in polar form, you have
$$c = r e^{i \theta}  ,$$
where $c \in \mathbb{C}$ is any complex number, $r \in \mathbb{R}$ is its distance from the origin, and $\theta \in [0,2\pi)$ is its phase.
So now, you can first find the first root, i.e. $w_1$, in a "regular way":
$$w_1 = \sqrt[4]{81} = 3.$$
Note that this is the root located on $x$ axis, so its angle $\theta$ in polar form is $0$, which is why $w_1 = r e^{i \theta} $ lets you determine the $r$ distance (since $e^{i \cdot 0} = 1$). Thus you have that $r=3$
Next, since you know that you are taking the fourth root, by the fundamental theorem of algebra, you know that you must have $4$ roots. Then, taking a shortcut, we have $\Delta \theta = 2 \pi /4 = \pi/2$. (Or you can also write it in a more formal way with De Moivre's theorem in polar form.) Then, your other three roots would be:
$$w_2 = re^{i \pi/2}=3 e^{i \pi/2} = ...$$
$$w_3 = re^{i 2\pi/2}=3 e^{i 2\pi/2} = ...$$
$$w_4 = re^{i 3\pi/2}=3 e^{i 3\pi/2} = ...$$
Compute these answers; for each of the cases then substitute into $ z^2-z -w=0$ and proceed to computing $z$ for each case (with the computation of the determinant and using De Moivre or the shortcut described above).
