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I'm trying to work through the following problem:

Prove that if the holomorphic function $f$ has an isolated singularity at $z_{0}$, then the principal part of the Laurent series of $f$ at $z_{0}$ converges in $\mathbb{C}\setminus\{z_{0}\}$.

My Thoughts: If $f$ has the Laurent series expansion $f(z)=\sum_{n=-\infty}^{\infty}a_{n}(z-z_{0})^{n}$, then the principal part of the Laurent series is $\sum_{n=-\infty}^{-1}a_{n}(z-z_{0})^{n}$. I'm assuming that I need to break this up to consider all possible types of singularities that could occur at $z_{0}$ (i.e., a removable singularity, a pole, or an essential singularity).

  • If $z_{0}$ is a removable singularity, the principal part of the Laurent series is trivial ($a_{n}=0$ for all $n<0$), so there is nothing to show.

  • If $z_{0}$ is a pole (say of order $k$), then $a_{-k}\neq 0$, but $a_{n}=0$ for all $n<-k$. Then the principal part is $\sum_{n=-k}^{-1}a_{n}(z-z_{0})^{n}$.

  • If $z_{0}$ is an essential singularity, then the principal part of the Laurent series has infinitely many non-vanishing terms. Then the principal part is $\sum_{n=-\infty}^{-1}a_{n}(z-z_{0})^{n}$.

My Questions: Am I right in saying that the convergence is trivial in the case that $z_{0}$ is a removable singularity? Also, how would I go about the case where $z_{0}$ is a pole or an essential singularity? Is it some sort of Cauchy-Hadamard argument?

Thanks in advance for any suggestions.

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