I'm new to complex analysis and am not sure where to start with this. The question states:
Let the origin be a pole of order m > 0 of an otherwise analytic function f of a complex variable. Show that f / f' has a removable singularity at z=0.
Would appreciate any suggestions!
Best Answer
Since $f$ is otherwise analytic, we can expand $f$ as a Laurent series
$$f(z)=\sum_{k=-m}^\infty c_kz^k,\;\;\;\;f'(z)=\sum_{k=-m}^\infty kc_kz^{k-1}$$
The pole is of order $m$, so $c_{-m}\neq 0$. This gives us
$$\frac{f(z)}{f'(z)}=\frac{\sum_{k=-m}^\infty c_kz^k}{\sum_{k=-m}^\infty kc_kz^{k-1}}$$
We multiply the top and the bottom by $z^{m+1}$ to get
$$\frac{f(z)}{f'(z)}=z\frac{\sum_{k=0}^\infty c_{k-m}z^k}{\sum_{k=0}^\infty(k-m)c_{k-m}z^k}$$
The fraction on the right is continuous at $0$, so its limit as $z\to 0$ is $-\frac{1}{m}$. Since this limit exists, the limit of the whole function is $0$, since the first piece (namely, $z$) goes to $0$.
I hope this helps!