Let $f(z)$ be analytic in the disk $|z|
All my attempts failed. I wanted to use the Laurent series at $z_0$ but the problem needs expansion at $0$.
Let $f(z)$ be analytic in the disk $|z|
All my attempts failed. I wanted to use the Laurent series at $z_0$ but the problem needs expansion at $0$.
There is an $a\ne0$ such that the principal part of $f$ at $z_0$ is given by $z\mapsto{a\over z-z_0}$. The function $g(z):=f(z)-{a\over z-z_0}$ has a removable singularity at $z_0$ and is analytic otherwise in $D_R$, which implies that $g$ is in fact analytic in $D_R$. So $g$ has a Taylor expansion $g(z)=\sum_{n=0}^\infty b_nz^n$ which is convergent for $|z|
Consider $\displaystyle f(z) = \frac{g(z)}{z-z_0}$. Since $f$ has a simple pole at $z_0$, $g$ is holomorphic on the disc.
Now, let $f = \sum a_nz^n, g=\sum b_nz^n$. Then, $\sum b_nz^n = (z-z_0)\sum a_nz^n$. Comparing the coefficients, you get $b_{n+1} = a_n - z_0a_{n+1}$.
As $n \to \infty$, $b_n \to 0$ while $a_n$ will not tend to $0$.