# $f$ has a pole at $z=a$ implies $1/f$ has a removable singularity at $z=a$

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.

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.