I'm trying to find the domain on which a function is analytic, specifically, $\text{Log}\left(\frac{1}{z}+i\right)$.
Would I need to find $\text{Log}\left(\frac{1}{z}+i\right)=\ln(r)+i\theta$ to find the branch cut?
How would I do that if $z=\frac{1}{z}+i$?
The branch points of the logarithm function, $\log(w)$, are at $w=0$ and the point at infinity.
If $w=\frac1z+i$, then $w=0$ when $z=i$. And the point at infinity in the $w$-plane maps to $z=0$ in the $z$-plane.
The principal branch in the $w$-plane lies along the negative real axis. This branch cut maps to the contour defined parametrically $z=\frac1{t-i}$, $t\in (-\infty,0]$, in the $z$-plane.
This contour is the semi-circular arc $x^2+(y-1/2)^2=1/4$, $x\in [-1/2,0]$. And this is the branch cut for the principal branch of the function $\log\left(\frac1z+i\right)$.