Originally Posted by

**tonio** As $\displaystyle z=-1$ is NOT a solution, we can do:

$\displaystyle (z-1)^4+(z+1)^4=0\Longleftrightarrow \frac{(z-1)^4}{(z+1)^4}=-1\Longleftrightarrow \left(1 - \frac{2}{z+1}\right)^4=-1$.

Solve now $\displaystyle w^4=-1=e^{\pi i+2k\pi i}=cis(\pi+2k\pi)\,,\,k=0,1,2,3\Longrightarrow w=e^{\frac{\pi i}{4}\left(2k+1\right)}$ , using De Moivre's Theorem, and once you get the roots

just substitute $\displaystyle w = 1-\frac{2}{z+1}$ and solve for $\displaystyle z$ .

If you don't understand something of the above then this question isn't for you, yet. Complex numbers enter usually in 11th or 12th years (5ht or 6th high school).

Tonio