How to find complex solutions of equations?

Question

I need help solving this task, if anyone had a similar problem, it would help me.

The task is: Determine all solutions of the equation in the set of complex numbers.

$(1-z)^5=z^5$

I thought I could include it $z=x+iy$, but I see that it leads nowhere.

Thanks in advance !

Answer 1

$0$ is not a root of the equation.

Now, let $\frac{1}{z}-1=w.$

Thus, we need to solve: $$w^5=1.$$ Can you end it now?

Answer 2

Hint:

We have $$\left({\frac{z}{1-z}}\right)^5=1=cos(2k\pi)+isin(2k\pi)$$

we now apply De Moivre's theorem

$$\frac{z}{1-z}=cos(2k\pi/5)+isin(2k\pi/5)$$

Answer 3

There is also the following way: $$z^5-(1-z)^5=(2z-1)(z^4-2z^3+4z^2-3z+1)=$$ $$=(2z-1)\left(\left(z^2-z+\frac{3}{2}\right)^2-\frac{5}{4}\right)=$$ $$=(2z-1)\left(z^2-z+\frac{3-\sqrt5}{2}\right)\left(z^2-z+\frac{3+\sqrt5}{2}\right).$$