Find all analytic functions in the disk $|z-2|<2$ such that $f \left(2+\frac{1}{n} \right) = \frac{1-n}{5n+3} + \sin{\left(\frac{n \pi}{2} \right)}$ Find all analytic functions (or prove that no such exist) inside the disk $|z-2|<2$ that satisfy the following condition:
$$f \left(2+\frac{1}{n} \right) = \frac{1-n}{5n+3} + \sin{\left(\frac{n \pi}{2} \right)}, \ n \in \mathbb{N}$$
For $n=2k$ the expression simplifies quite a bit since $\sin{\left(\frac{n \pi}{2} \right)}= \sin{\left(k \pi \right)} = 0$ so we're left with $$f \left(\frac{4k+1}{2k}\right) = \frac{1-2k}{10k+3}$$
Similarly for $n=4k-3$ and $n=4n-1$ we can simplify the sinus expression to $1$ and $-1$. So we're getting $3$ different expressions for $f\left(2+\frac{1}{n}\right)$ based on whether $n=2k$ , $n=4k-3$ or $n=4k-1$. My idea is to use the identity theorem for any of these two expressions and prove that no such function exists but I can't quite tie it all together.

If $n$ is a multiple of $4$, then\begin{align}f\left(2+\frac1n\right)&=\frac{1-n}{5n+3}\\&=\frac{1/n-1}{5+3/n}\\&=\frac{2+1/n-3}{-1+3(2+1/n)}.\end{align}Now, for each $z\in D_2(2)$, let $g(z)=\frac{z-3}{-1+3z}$. Then, by the identity theorem, $g=f$. But that's impossible, since $g(3)\ne f(3)$: $g(3)=0$, whereas$$f(3)=f\left(2+\frac11\right)=1.$$So, no such function $f$ exists.
