How many roots does the equation $z^{2018}=2018^{2018}+i$ have?

Question

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.

Answer 1

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

$$r^{2018}\sin(2018\theta)=1$$

Answer 2

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 $\forall k\in[0,2017]$

$$\large z=r^{1/2018}e^{i\cdot\frac{\theta+2k\pi}{n}}$$

Answer 3

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.

Answer 4

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.