In Section V.1 of Conway's Functions of One Complex variable, he says that if $f$ has a pole at $z=a$ implies $[f(z)]^{-1}$ has a removable singularity at $z=a$. I am confused why $[f(z)]^{-1}$ should have an isolated singularity at $z=a$ in the first place.
For example, take $f(z) = 1/z $. Then, $[f(z)]^{-1} = z$. Here, $f$ has a pole at $z=0$ whereas $[f(z)]^{-1}$ is entire and has no singularities.
Best Answer
Watch the domain. In your example, $f(z)=\frac1z$ for all $z\neq 0$. The reciprocal $g(z)=\frac1{f(z)}$ is then equal to $z$ for all $z\neq 0$. At zero? It's not defined. That's a classic removable singularity.