Problem: Show that there is no function $f$, holomorphic on a neighbourhood of $z=0$, such that its value on $z=\frac{1}{n}$ is $(-1)^n (\frac{1}{n})$.
To contrast, find a function holomorphic around $z=0$ such that $f(\frac{1}{n})=\frac{n}{n+1}$.
I am stumped, it's not that I do not know the course material, I just don't know where to start. What would a good starting point be?
These problems were assigned along with a problem set on applications of Cauchy's Integral formula and Poles and Residues.
Thank you for your time.
For the first problem, consider the sequences $a_n = 1/n$ for $n$ even and $n$ odd separately. Use the fact that the functions $f + z$ and $f - z$ would both have zero sets with accumulation points.
More generally, a good starting point for these is to find a related holomorphic function whose set of zeros has an accumulation point in the interior of the domain - such a function must then be identically zero.
For the second problem, note that you can rearrange the fraction as
\begin{align*} f\left(\color{red}{\frac 1 n}\right) = \frac{n}{n + 1} &= \frac{1}{1 + \color{red}{1/n}} \end{align*}
This should strongly suggest what $f$ must be.