Let $f$ be an analytic function in the annulus $0 < |z| < 1 $ such that it's singularity in $z=0$ is not essential. I want to find all of such functions $f$ that satisfy for $n = 3, 4,...$:
- $f(\frac 1n)=\frac {n^4}{1+n}$
- $f(\frac 1n)=\frac {\sqrt{n^2-4}}{2n}$
- $f(\frac 1n)=\frac 1{2^n}$
I'm pretty clueless on how to even start. I thought using the fact that $z^mf(z)$ is analytic in the unit disk, and then try to satisfy the conditions, but I still couldnt understand how to do that. Could you please give me a hint?
1. As you said there exists a $m$ so that $g(z)=z^mf(z)$ is Analytic. Moreover, as $f$ is not continuous at $z=0$, the singularity is not removable, hence it must be a pole. Therefore, we can pick the $m$ in such a way that $g(0) \neq 0$.
As $$g(\frac{1}{n})=\frac{1}{n^m}\frac {n^4}{1+n}$$ and $\lim_n g(\frac{1}{n}) =g(0) \neq 0$ it follows that $m=3$ and
$$g(\frac{1}{n})=\frac {n}{1+n}=\frac{1}{1+\frac{1}{n}}$$
Therefore the Analityc functions $g(z)$ and $\frac{1}{1+z}$ are equal on the sequence $\frac{1}{n}$.
2. Hint $f(\frac{1}{n})=\frac{\sqrt{1-\frac{4}{n^2}}}{2}$ has a finite limit at $z=0$. Deduce that the singularity is removable (same way as in 1: if there is a pole, by multiplying by the right $z^m$ you get an Analytic function with a non-zero limit).
3. Same as $2$, prove that $f$ has a removable singularity at $z=0$, and that $f(0)=0$.
Let $m$ be the order of the zero at $z=0$. What can you say then about $$\lim_{z \to 0} \frac{f(z)}{z^m}$$
what happens when you calculate the limit along $z =\frac{1}{n}$?