How to solve the complex root $z^\sqrt5 =1$ with square root? 
Solve 
  $$z^\sqrt5 =1$$
  for $z$ and state how many unique solutions are possible.

I tried to convert $1$ to polar form and got $z=\exp^\left((2k\pi+2\pi)i/\sqrt5\right)$. Could someone please help me out?

Let $z=re^{i\theta}$.
Since $z^\sqrt5=r^\sqrt5e^{i\sqrt5\theta}=1=1e^{\left(0+2k\pi\right)i}$ for $k \in \mathbb{Z}$, $r=1$.
Now we have $e^{i\sqrt5\theta}=e^{2k\pi i}$, which is exactly where you have reached.
Then we have $\sqrt5\theta=2k\pi$
So according to you, you are using $0 \leq \theta \leq 2\pi$. 
We know from earlier that $\theta=\frac{2k\pi}{\sqrt5}$.
So here is the list of the values of $\frac{2k}{\sqrt5}$ corresponding to some $k$'s.
\begin{align}
&k=-1, &&\frac{2k}{\sqrt5}=-0.894427191 \\
&k=0, &&\frac{2k}{\sqrt5}=0 \\
&k=1, &&\frac{2k}{\sqrt5}=0.894427191 \\
&k=2, &&\frac{2k}{\sqrt5}=1.788854382 \\
&k=3, &&\frac{2k}{\sqrt5}=2.683281573 \\
\end{align}
Which values of them are within the range now?
