$\textit{Find all solutions of the given equation:}$ $$(z+1)^4 = 1 - i.$$
$1 - i$ on the xy plane is equivalent to $(1, -1)$, which has an angle $\theta = \frac{-\pi}{4}$ and length $\sqrt{2} = 2^{1/2}$.
In polar representation, $$1-i = 2^{1/2} e^{i(- \pi/4 + 2k \pi)} \quad k \in \mathbb{Z}.$$
Back to the original question,
\begin{align*}
(z+1)^4 &= 2^{1/2} e^{i(- \pi/4 + 2k \pi)} \\
z+1 &= (2^{1/2} e^{i(- \pi/4 + 2k \pi)})^{1/4} \\
z &= 2^{1/8} e^{i(- \frac{\pi}{16} + \frac{1}{2}k \pi)} - 1
\end{align*}
We only need to solve for $k = 0, 1, 2, 3$ since $(z+1)^4$ has 4 roots.
$$
z =
\begin{cases}
2^{1/8} e^{i \left(- \frac{\pi}{16} \right)} - 1 = 2^{1/8} (\cos\frac{-\pi}{16} - i \sin\frac{-\pi}{16}) & \text{for } k=0, \\
2^{1/8} e^{i \left(\frac{7\pi}{16} \right)} - 1 = 2^{1/8} (\cos\frac{7\pi}{16} - i \sin\frac{7\pi}{16}) & \text{for } k=1, \\
2^{1/8} e^{i \left(\frac{15\pi}{16} \right)} - 1 = 2^{1/8} (\cos\frac{15\pi}{16} - i \sin\frac{15\pi}{16}) & \text{for } k=2, \\
2^{1/8} e^{i \left(\frac{23\pi}{16} \right)} - 1 = 2^{1/8} (\cos\frac{23\pi}{16} - i \sin\frac{23\pi}{16}) & \text{for } k=3
\end{cases}
$$
