Complex numbers such that $(z-4)^4=-1$ 
Find all complex numbers $z$ in Cartesian form such that $(z-4)^4=-1$.

This is what I did: I squared root both side making $-1$ into $i$ and $(z-4)^2$ and I took square of both side again and had $i^2$ which is equal to $-1$ and the other side $(z-1)$. At the end I had $z=0$.


This is what I did: I squared root both side making $-1$ into $i$ and $(z-4)^2$

This is on the right track. Except, when you square root both sides, you need to do $\pm$. So instead of getting $i$ you get $\pm i$:
$$
(z - 4)^2 = \pm i
$$

and I took square of both side again and had $i^2$ which is equal to $-1$ and the other side $(z-1)$. At the end I had $z=0$.

The next step is to take the square root again (not squaring both sides), is that what you meant to do?
This part is a little tricky: when we take square root of both sides we need to know what are all the square roots of $i$ and $-i$.
The square roots of $i$ are
$$
\pm \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i \right)
$$
and the square roots of $-i$ are
$$
\pm \left( \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i \right)
$$
So, when we take the square root of the equation $(z - 4)^2 = \pm i$, we get four possibilities (the two square roots of $i$, and the two square roots of $-i$):
\begin{align*}
z - 4 &= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i \\
z - 4 &= \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i \\
z - 4 &= -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i \\
z - 4 &= -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i.
\end{align*}
Can you find the answer in all of the four cases?
