$f$ has a pole at $z=a$ implies $1/f$ has a removable singularity at $z=a$
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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.
complex-analysis analytic-functions singularity
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asked Jan 6 at 10:01
Ajay Kumar Nair Ajay Kumar Nair
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