let $f : U \rightarrow U$ be an analytic function, whose Taylor series at $0$ is $f(z) = z + a_2z^2 + a_3z^3 + ...$

Question

Let $U \subset \mathbb{C}$ be a bounded open set containing $0,$ and let $f : U \rightarrow U$ be an analytic function, whose Taylor series at $0$ is $f(z) = z + a_2z^2 + a_3z^3 + ...$

Prove that $a_2 = 0$.

Hint: Consider the functions $g_n(z) = f\circ ... \circ f(z)$.

I tried doing this and then noticed the $z^2$ term gets really big (the leading coefficient that is) but that didn't seem to do anything. Any suggestions?

Answer 1

You pretty much solved it already: $g_n(z)=z+na_2z^2 + \ldots$, so $g''_n(0)=2na_2$, but by Cauchy's formula this quantity is bounded.

More explicitly, if $\overline{D}_R(0)\subset U$, then $$ g''_n(0) = \frac{1}{i\pi} \int_{\partial D_R} \frac{g_n(w)}{w^3}\, dw , $$ and this is $\lesssim 1/R^2$ since, by assumption, $g_n(w)$ takes values in the bounded set $U$.