$f(z)$ is analytic on unit disk $ \cup \{1\}, f(1)=1$ , then $f'(1)\geq0$ 
$D= \big\{z\in \Bbb C:|z| \lt 1\big\}$.  A function $f(z)$ such that 
1) $f(D)\subset D, f(1)=1$
2) $f(z)$ is analytic on $D \cup \{1\}$ 
show that $f'(1)\geq0$.

If $f'(1)\ne 0$, Could we get that $\arg(f'(1))=0 $? 
 A: Let's first prove the following lemma: assume $u,v$ real s.t $\Re{((u+iv)(z-1)+O((z-1))^2}) > 0$ for all $|z| <1$ in a small neighborhood of $1$, then $u \le 0,v=0$. 
Proof: picking $z=r<1$ but close enough to $1$, we get $u(r-1)+O((r-1)^2) >0$ hence $u \le 0$. If $v \ne 0$, pick $z=1-\epsilon^2\pm i\epsilon$ and for small enough $\epsilon>0$, $|z|<1$ so $z$ is in the given neighbourhood intersected with the open unit disc, $|z-1|^2=O(\epsilon^2)$ and $\Re{((u+iv)(z-1))+O((z-1))^2}=|u|\epsilon^2 \pm v\epsilon+O(\epsilon^2)=\epsilon (|u|\epsilon \pm v+O(\epsilon))$ which can be clearly made negative if we choose the right sign in $\pm$ and small enough $\epsilon >0$, so contradiction.
With this lemma let's prove the result. Take $g=\frac{1-f}{1+f}$, $g$ is analytic on the unit disc, extends analytically at $1$, and $\Re g >0$ on the unit disc, while $g(1)=0$. Plainly $2g'(1)=-f'(1)$ so we need to prove that $g'(1) \le 0$. We can write the Taylor series on a small disc around $1$ as $g(z)=(u+iv)(z-1)+O((z-1)^2)$ and $\Re g >0$ on the piece of that disc that belongs to the unit disc. Applying the lemma we get $u \le 0, v=0$ and obviously $g'(1)=u \le 0$, so done!
A: $u(z) = \operatorname{Re}f(z)$ is defined (and differentiable) in a neighborhood of $z=1$, with $u(z) \le 1$ for $|z|\le 1$ and $u(1) = 1$. It follows that
$$
 h(t) = u(e^{it})
$$
has a maximum at $t=0$, so that
$$
 0 = h'(0) = u_x(1) \sin(0) -i u_y(1) \cos(0) = -iu_y(1) \, ,
$$
i.e. $u_y(1) = 0$. Also $u(x) < 1$ for $0 \le x < 1$, so that
$$
 u_x(1) = \lim_{x \to 1^-} \frac{1-u(x)}{1-x} \ge 0 \, .
$$
Combining these results we get
$$
 f'(1) = u_x(1) + iv_x(1) = u_x(1) -i u_y(1) = u_x(1) \ge 0 \, .
$$
