Let $f$ be a function holomorphic in $B(0,1)\cup \left\{ 1 \right\}$ with $$f(B(0,1))\subset B(0,1),\ \ f(1)=1$$ Prove that $f'(1)\ge 0$.
My attempt: I have tried to find some property of $f$ when $f'(1)\neq 0$, and guessed if I can prove $\operatorname{arg}f'(1)=0$, but don't know how to proceed specifically. Thanks for any help.
Here is a proof based only on the existence of derivative in the sense that $$ f'(1) = \lim_{z\to 1,\ z\in B(0,1)}\frac{f(z)-f(1)}{z-1} $$ exists. For every $\theta \in (-\pi/2,\pi/2)$ the point $1-re^{i\theta}$ lies in $B(0,1)$ when $r>0$ is small enough. By the definition of derivative, $$ f(1-re^{i\theta}) = 1 - f'(1) re^{i\theta} + o(r),\qquad r\to 0 $$ Since $|f|<1$, the real part of $1 - f'(1) re^{i\theta}$ cannot exceed $1$; this implies $\operatorname{Re}(f'(1) e^{i\theta})\ge 0$. Writing $f'(1)$ in polar coordinates, $f'(1) = \rho e^{it}$, makes it clear that $t=0$, for otherwise $\exp(i(t+\theta))$ has negative real part for some $\theta\in (-\pi/2,\pi/2)$.