I really don't know how to solve this problem!
Consider $F$, an analytic fuction, so that,
$$f(z)=F(\frac{1}{z-1})$$
has a pole. Demonstrate that F(z) is a polynomial
So, I tried to make a variables change, and applied the Residue theorem, but I couldn't make it!
What would you do?
I think that $F$ is entire (F Analytic over $\mathbb{C})$. Otherwise it's not true.
If $F$ is entire : f has at most one singularity at 1. The exercice says that 1 is a pole. So f is a meromrphic function having a single pole at 1. let $ \frac{a{-n}}{(z-1)^n}+\cdots\frac{a_{-1}}{z-1}+a_0+a_1(z-1) \cdots $ be the Laurent series of $f$ at 1. and $P(z)=a_{-1}z+\cdots a_{-n}z^n$. $g(z)=f(z)-P(\frac{1}{z-1})$ is holomorphic over $\mathbb{C}$. On the other hand one check's that $g$ is bounded. Therefor by Liouvilles theorem $g(z)=a$ where a is a constant. Finaly we get $F(z)=P(z)+a$ (a=a_0).