Question: Solve $z^4 = 1$ for all $z$.
Hence or otherwise, solve $z^4 =(z-1)^4$.
My attempt for the first part
$$z^4 = 1$$
$$z^4 -1 =0$$
$$ (z^2-1)(z^2+1) = 0 $$
$$ (z-1)(z+1)(z+i)(z=i)=0$$
$$ z = 1,-1,i,-i$$
However I am stuck on the second part , what hint does the first part give to solve the second part. I'm referring to the "otherwise".
I think I get the second part now
$$z^4 =(z-1)^4$$
$$ z^4 - (z-1)^4 = 0 $$
Let p = z , q = (z-1)
$$ p^4 - q^4 = 0 $$
$$ (p^2-q^2)(p^2+q^2)=0 $$
$$ (p-q)(p+q)(p^2+q^2)=0 $$
$$ (z-(z-1))(z+(z-1))(z^2+(z-1)^2) = 0 $$
$$ (1)(2z-1)(z^2 + z^2 - 2z + 1) = 0 $$
$$ (2z-1)(2z^2 -2z +1)=0$$
$$ z = \frac{1}{2} , \frac{1+i}{2} , \frac{1-i}{2} $$
However am I suppose to get 4 solutions not 3?
You are completely correct on the first question so let's work on the second one.
We can divide both sides by $(z-1)^4$. We now have to find the solutions to the equation ${z^4 \over(z-1)^4} = 1$ or $({z\over z-1})^4 = 1$. Take the 4th root of both sides and you get $1^{1/4} = {z \over (z-1)}$. There are a couple of possible answers for $1^{1/4}$. $1, -1, i, -i$ specifically. Now all you have to do is find the specific solutions for each of the branches. I'll leave the algebra to you as I'm pretty sure you can do it from here.
One of the roots above doesn't end up working because of what we did to the domain. $1 = {z \over (z-1)}$ doesn't have a solution because it implies that $z-1 = z$ or that $-1 = 0$. We have to discard this root so we are left with three roots.