Solve in C :
$\displaystyle {z^4} + {\left( {1 + z} \right)^4} = 0 \Leftrightarrow$
$\displaystyle \Leftrightarrow {z^4} + 2{z^2}{\left( {1 + z} \right)^2} + {\left( {1 + z} \right)^4} - 2{z^2}{\left( {1 + z} \right)^2} = 0 \Leftrightarrow$
$\displaystyle \Leftrightarrow {\left( {{z^2} + {{\left( {1 + z} \right)}^2}} \right)^2} - 2{z^2}{\left( {1 + z} \right)^2} = 0 \Leftrightarrow $
$\displaystyle \Leftrightarrow \left( {{z^2} + {{\left( {1 + z} \right)}^2} - \sqrt 2 z\left( {1 + z} \right)} \right)\left( {{z^2} + {{\left( {1 + z} \right)}^2} + \sqrt 2 z\left( {1 + z} \right)} \right) = 0.$
What you need to do now, I hope you know
Another method: $\displaystyle z^4 + (1 + z)^4 = 0\: \Leftrightarrow\: (1+z)^4 = (-1)z^4\: \Leftrightarrow\: 1+z = \omega z \: \Leftrightarrow\: z = \frac1{\omega-1}$, where $\displaystyle \omega$ is one of the fourth roots of –1, namely $\displaystyle \tfrac1{\sqrt2}(\pm1\pm i)$.