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}