How many roots does the equation $z^{2018}=2018^{2018}+i$ have? 
Consider the equation
  $$ z^{2018}=2018^{2018}+i$$ where $i=\sqrt{-1}$.
  How many complex solutions as well as real solutions does this equation have?

My attempt:
I took the polar form as the equation has very difficult to handle when using $z=x+iy$.
So I set $z=re^{iθ}$, which yields
$$ (re^{iθ})^{2018}=2018^{2018}+e^{i\frac{\pi}{2}}$$
After this I was not able to handle it. 
 A: hint
If $$z=re^{i\theta}$$ then
$$z^n=r^n(\cos(n\theta)+i\sin(n\theta))$$
the real part gives
$$r^{2018}\cos(2018\theta)=2018^{2018}$$
and the imaginary
$$r^{2018}\sin(2018\theta)=1$$
A: HINT
We have that
$$r=|2018^{2018} + i|=\sqrt{2018^{4036}+1}$$
$$\theta =Arg(2018^{2018} + i)=\arctan \left(\frac1{2018^{2018}}\right)$$
then
$$2018^{2018} + i=\sqrt{2018^{4036}+1}\,e^{i\theta}$$
now use $\forall k\in[0,2017]$
$$\large z=r^{1/2018}e^{i\cdot\frac{\theta+2k\pi}{n}}$$
A: There are several red herrings in the question and, as the problem is stated, you don't need to describe the solutions.
Your problem can be generalized to $z^n=a+i$, where $n$ is a positive integer and $a$ is real. Clearly, $z$ cannot be real, nor can $a+i$ be zero.
Thus the solutions are all complex (not real) and they are the $n$-th roots of $a+i$. There are $n$ of them.
A: The equation $z^{2018} = 2018^{2018} + i$ has a total of 2018 roots, in which some are real. It is clear that there is no real roots. So it has 2018 complex roots.
