I have to find the solution of the following complex equation: $z^5+(1+i)z=0$, so first of all I did $z(z^4+1+i)=0$, which gives me $z=0$.
Now, all I have to do is to solve $z^4+1+i=0$, so I did $z^4=-1-i$ and using De Moivre's formula, I get $z^4=1(\cos(4\alpha)+i\sin(4\alpha))$.
Then, solving the system of equations $\begin{cases}\alpha=\frac{\pi k}{2}\\\alpha = \pi+2k\pi\end{cases}$, it gives me that $k=-\frac{9}{12}$, which is impossible, since $k\in\mathbf{Z}$, yet using Mathematica it gives me 5 solutions. I know that I should get 5 solutions, but I can't understand where I did wrong.
For nonzero roots $z^4=\sqrt{2}\exp(-\pi i+2n\pi i)$, so $z=\sqrt[8]{2}\exp\left(-\frac{\pi i}{4}+\frac{n\pi i}{2}\right)$ with $n\in\{0,\,1,\,2,\,3\}$. As far as I can tell, you overlooked the modulus and tried to set a sine and cosine of the same angle to $-1$, in contradiction of the Pythagorean theorem.