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If $f$ is meromorphic on $\mathbb{C}$ and $a \in \mathbb{C}$, why is $g=\displaystyle \frac{1}{f-a}$ holomorphic (analytic), even if there are poles?

Thank you very much for your time,


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up vote 2 down vote accepted

The function $g(z)$ doesn't have to be holomorphic in $\mathbb{C}$, because if there are $z\in\mathbb{C}$ such that $f(z)=a$, then $g$ will have poles, and hence wont be holomorphic in $\mathbb{C}$, but it would always be holomorphic in $\{z:f(z)\neq a\}$, and clearly meromorphic in $\mathbb{C}$. If, for some reason, you know that $f(z)\neq a$ for all $z$, then $g(z)$ will be holomorphic in $\mathbb{C}$.

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Sorry, my questions was very badly phrased. I've edited. Thank you for your input and sorry again! – Chris Jun 17 '12 at 13:10
You are absolutely right, $f(z)-a$ has no zeroes (this is part of the proof of Picards theorem for meromorphic functions). – Chris Jun 17 '12 at 13:29

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