Question: I have tried to solve the following equation:

$$
(z+1)^4+4(z-1)^4=0 \quad,\quad z=(a+bi)
$$

If I could leave the $z^4$ alone I could solve it using the formula, but I do not know how to operate with it having additions and substractions. 

Can someone please help me?
Thank you!

Answer: 
Another idea:
$$(z+1)^4 = - 4(z-1)^4$$
iff
$$z+1 = \root2\of2(z-1)\omega$$
with $\omega =$ some 4th root of $-1$ and
$$z = \frac{\sqrt2\omega+1}{\sqrt2\omega-1}$$
for the different values of $\omega$.
Can you factor $(z+1)^4 + 4(z-1)^4$ (or perhaps an equivalent form of this) into the product of two quadratics? (Hint: Yes)