Can someone help me with this question. Or just can give a hint which theorem from complex Analysis to use to solve this. Really stucked!
Q. Let f(z) = $\sum_{n\ge0} a_{n}z^{n}$ be an analytic function on the open unit disc D around $0$ with $a_{1}\neq0$. Suppose that $\sum_{n\ge2} |na_{n}|\le a_{1}$. Then of the following are true.
A) There are only finitely many such $f$.
B) $|f'(z)|\gt 0$ for all $z$.
C) If $z$, $w$ $\in D$ are such that $z\neq w$ and $f(z) =f(w)$, then $a_{1}$ = $\sum_{n\ge2}a_{n}(z^{n-1} + z^{n-2}w +...+ w^{n-1})$.
For part B) Its true because as given $a_{1}\neq 0$ hence $f'(z) \neq 0$ therefore $|f'(z)|\gt 0$ for all $z$.
For part C. I am not sure where to start..
A): $f(z)=a_1z$ satisfies the hypothesis for every $a_1 >0$ so A) is false.
Suppose $f'(z)=0$. Note that $f'(0)=a_1>0$ so $z \neq 0$. We have $-a_1z=\sum_{n \geq 2} na_nz^{n-1}$ so triangle inequality (after division by $z$) gives $|a_1|\leq \sum_{n \geq 2} |na_n|\leq a_1$ with strict inequality unless $a_n=0$ for all $n \geq 2$. But if $a_n=0$ for all $n \geq 2$ then $f'(z)=a_1\neq 0$, a contradiction. Thus, B) is true.
For C) just use the fact that $(z-w)(z^{n-1} + z^{n-2}w +...+ w^{n-1})=z^{n}-w^{n}$