Let \(U\subset\CC\) be a bounded open set containing \(0\), and let \(f\colon U\to U\) be an analytic function, whose Taylor series at \(0\) is
\[ f(z)=z+a_2z^2+a_3z^3+\cdots \]
Prove that \(a_2=0\). (Hint: consider the functions \(g_n(z)=f\circ\cdots\circ f(z)\) obtained by composing \(f\) with itself \(n\) times.)
Solution
For \(n\ge1\), let \(g_n=f^{\circ n}\). Choose \(R>0\) such that \(|w|\le R\) for every \(w\in U\), and choose \(r>0\) such that \(\{z:|z|\le r\}\subset U\).
1. For every \(n\ge1\),\[ g_n(z)=z+na_2z^2+O(z^3) \]as \(z\to0\).
For \(n=1\) this is the given Taylor expansion of \(f\). Suppose \[ g_n(z)=z+na_2z^2+O(z^3). \] Then \(g_n(z)=z+O(z^2)\), so \[ g_n(z)^2=z^2+O(z^3) \quad\text{and}\quad g_n(z)^3=O(z^3). \] Using \[ f(w)=w+a_2w^2+O(w^3) \] with \(w=g_n(z)\) gives \[ \begin{aligned} g_{n+1}(z) &=f(g_n(z))\\ &=g_n(z)+a_2g_n(z)^2+O(g_n(z)^3)\\ &=z+(n+1)a_2z^2+O(z^3). \end{aligned} \] The claim follows by induction.
2. For every \(n\ge1\) and every \(z\) with \(|z|=r\),\[ |g_n(z)|\le R. \]
Each iterate \(g_n\) maps \(U\) into \(U\) because \(f(U)\subset U\). Since the circle \(|z|=r\) lies in \(U\), one has \(g_n(z)\in U\), and the choice of \(R\) gives the estimate.
3. For every \(n\ge1\),\[ n|a_2|\le\frac{R}{r^2}. \]
By step 1, the coefficient of \(z^2\) in the Taylor expansion of \(g_n\) at \(0\) is \(na_2\). Cauchy’s coefficient formula therefore gives \[ na_2 = \frac{1}{2\pi i} \int_{|z|=r}\frac{g_n(z)}{z^3}\,dz. \] Using step 2 and the fact that the circle has length \(2\pi r\), \[ \begin{aligned} n|a_2| &\le \frac{1}{2\pi}(2\pi r)\frac{R}{r^3}\\ &= \frac{R}{r^2}. \end{aligned} \]
4. One has\[ \boxed{a_2=0}. \]
Step 3 yields \[ |a_2|\le\frac{R}{nr^2} \] for every \(n\ge1\). Letting \(n\to\infty\) gives \(|a_2|=0\).
5. Q.E.D.
Step 4 is the required conclusion.