Solving complex equation, $z^4 = (z + 1)^4$ We just started learning about complex numbers, and I'm having some trouble on how to solve this problem. 
$$z^4 = (z + 1)^4$$
Was wondering if anyone could help me out.
 A: $(z+1)^4-z^4=((z+1)^2+z^2)((z+1)^2-z^2)=(2z^2+2z+1)(2z+1)$, which gives the answer:
$$\left\{-\frac{1}{2}-\frac{1}{2}i,-\frac{1}{2}+\frac{1}{2}i,-\frac{1}{2}\right\}$$
A: Hint: $z^4=(z+1)^4$ i.e $z=e^{2i\pi\over 4}(z+1), i=0,1,2,3$.
A: HINT:
There is one (real) solution that you can find by inspection:

 Since $z^4 = (z+1)^4$ is kind of "symmetric" with respect to $z=-1/2$, it makes sense to evaluate this number. You will find it is a solution for the equation.

Once you find it, can you see a way of transforming this equation into a degree 3 polynomial equation? if so, and since you already know one solution, it should be easy (applying synthetic division and the quadratic formula) to find the other two roots.
A: the equation implies
$$
|z|^2=|z+1|^2
$$
so
$$
z\bar z = z\bar z+z+\bar z +1
$$
giving $$\mathfrak{R}(z) = -\frac12$$
set $z=-\frac12 + iy$ ($y \in \mathbb{R}$)
so
$$
(-\frac12 + iy)^4 = (\frac12 + iy)^4
$$
giving (since the real terms on either side cancel)
$$
4(-\frac12)^3iy + 4(-\frac12)(iy)^3 = 4(\frac12)^3iy + 4(\frac12)(iy)^3 
$$
i.e.
$$
 y-4y^3=0
$$
so the three roots of the original cubic are
$$
z= -\frac12, -\frac12 \pm \frac{i}2
$$
A: Clearly $z=0$ is not solution. Thus you can divide by $z^{4}$. 
You get $((z+1)/z)^4 =1$.
You know the solutions to $X^4 = 1$ in the complex numbers are $\pm 1$ and $\pm i$. 
So you are down to solving  $(z+1)/z = 1$, $(z+1)/z=-1$, $(z+1)/z = i$, $(z+1)/z=-i$, which is easy. 
This is basically what Jack said in a comment. 
A: The equation implies  $z=t(z+1)$ with $|t|=1$ that means $z=\frac{t}{1-t}$ with $t\not=1$. Coming back to the equation we get $t^4=1$ that means $t\in\{-1,i,-i\}$ hence $z\in\{\frac{-1}{2},\frac{i}{1-i}, \frac{-i}{1+i}\}$
