Using complex analysis , how to prove that any holomorphic function $f:\overline{D(0;1)} \to \overline{D(0;1)}$ has a fixed point? Using complex analysis , how to prove that any holomorphic function $f:\overline{D(0;1)} \to \overline{D(0;1)}$ has a fixed point ? From this answer Suppose $f(z)$ is analytic in the closed unit disc... I can see that if $|f(z)|<1$ on $|z|=1$ then I can prove it easily ; but what if that condition doesn't hold ? I am stuck . Please help . Thanks in advance 

Only "using comple analysis".  
Take a sequence $(p_n)$ where $0<p_n<p_{n+1}<1$ and $p_n\to 1$ as $n\to \infty$. Let $f_n(z)=p_nf(z)$. Then $f_n(z)$ satisfies $|f_n(z)|<1$, so $f_n(z)$ has a fixed point $z_n$ $(|z_n|<1)$. Since $\overline{D(0;1)}$ is compact, there is a subsequence $(z_{n_k})$ which converges to some point $ z_0\in \overline{D(0;1)}$. Since $(f_n(z))$ converges uniformly to $f(z)$ on $\overline{D(0;1)}$,
$$f(z_0)-z_0=\lim_{k\to \infty} \left(f_{n_k}(z_{n_k})-z_{n_k}\right)=0.$$
Note: $z_0$ may lie on $|z|=1.$
Example: $f(z)=\frac{2z-1}{2-z}.$ This function satisfies $|f(z)|\le 1$ on $|z|\le 1$ and has two fixed points $z=1, -1$,  but $f$ has no fixed point in $|z|<1$.
