Find all solutions to:
1) $\displaystyle cosh(z)=-1$-----done
2) $\displaystyle cos^2 (z)=4$
3) $\displaystyle tan(z)=i$
Remember that
$\displaystyle \cos(z)=\frac{e^{iz}+e^{-iz}}{2}$
This gives
$\displaystyle \left(\frac{e^{iz}+e^{-iz}}{2} \right)^2=4 \iff e^{2iz}+2+e^{-2iz}=16$
$\displaystyle (e^{2iz})^2-14e^{2iz}+1=0$
This is quadratic in $\displaystyle e^{2iz}$
Can you finish from here?
You will use a similar trick for tangent just write it as exponentials.