Suppose $U\subset \mathbb{C}^n$ is a bounded domain and $f$ is a holomorphic map from $U$ to itself such that $f(a)=a$ for some $a\in U$. Prove that all eigen values $\lambda$ of $Df(a)$ lie in the unit disk.
I honestly have no idea how to start. Tried for $n=2$ for simplicity but dont know where to go from the complicated Jacobian. At least a hint is appreciated.
Since $U$ is bounded, the family $\{ f^k : k \in \mathbb{N}\}$ is a uniformly bounded family of holomorphic maps. By Cauchy's integral formula it follows that the families of partial derivatives
$$\left\{\frac{\partial(f^k)_i}{\partial z_j} : k \in \mathbb{N} \right\}$$
are locally uniformly bounded for all $i,j$. In particular, the components of $Df^k(a) = \bigl(Df(a)\bigr)^k$ are bounded independently of $k$. Therefore there is a constant $C$ such that $\lvert\mu\rvert \leqslant C$ whenever $\mu$ is an eigenvalue of some $Df^k(a)$. Since for an eigenvalue $\lambda$ of $Df(a)$, $\lambda^k$ is an eigenvalue of $Df^k(a)$, it follows that $\lvert\lambda\rvert^k \leqslant C$ for all $k$, and this implies $\lvert\lambda\rvert \leqslant 1$.
So all eigenvalues of $Df(a)$ lie in the closed unit disk. Of course eigenvalues of modulus $1$ are possible (e.g. for $f = \operatorname{id}_U$).