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,
Chris
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,
Chris
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 $\mathbb{C}$.