Finding all solutions $z=x+iy$ for four basic problems I am preparing for an exam on Complex Analysis and I have four small problems. Two of them I think I have solved but I would really appreciate a sanity check since I very often miss something. On the other two, I have no idea how to start, any hint/suggestion would be appreciated. I need to find all solutions $z=x+iy$.
$\textbf{a) } z+1 = \log{(i+1)}$. The typesetting won't allow it, but is is actually Log, hence the principal branch. 
$\begin{align}
z+1 &= \log{i+1}\\
z+1 &= \ln{\sqrt{2}} + \frac{\pi}{4}\\
z &= \ln{\sqrt{2}} + \frac{\pi}{4} -1
\end{align}
$
$\textbf{b) } \sin{z} = -i$. My solution is 
$\begin{align}
\frac{e^{iz}-e^{-iz}}{2i} &= -i\\
e^{iz}-e^{-iz} &= 2\\
e^{iz}-e^{-iz} -2 &= 0 \text{ , now let $a = e^{iz}$}\\
a^{2}-2a-1 &= 0\\
a &= 1 \pm \sqrt{2}
e^{iz} = 1 \pm \sqrt{2}\\
iz &= \ln{(1 \pm \sqrt{2})}\\
z &= -i\ln{(1 \pm \sqrt{2})}.
\end{align}
$
Now the two problems I do not know where to start:
$\textbf{c) } (z+i)^{3} = 8. $ Writing it out does not seem to help me get anywhere, I then find $z^{3}-3z+3iz^{2}-i=8$.
$\textbf{d) } (z+1)^{2i} =1$. No clue.

The first two are correct.
c) $(z+i)^3=8\iff z+i=2\vee z+i=2\left(-\frac12\pm\frac{\sqrt3}2i\right)$
d) I assume that $z^w$ is defined as $\exp\left(w\operatorname{Log}(z)\right)$. If so,\begin{align}(z+1)^{2i}=1&\iff\exp\left(2i\operatorname{Log}(z+1)\right)=1\\&\iff 2i\operatorname{Log}(z+1)=2\pi in\text{ (for some integer }n\text{)}\\&\iff\operatorname{Log}(z+1)=\pi n\\&\iff z+1=e^{\pi n}.\end{align}
