The topic comes from the last two exercises of Styhway, Liu Taishun, "complex function" 50 pages:
3. Set $f$ on $b (0,1) \cup\{1\}$ pure, and $ $f (b (0,1)) \subset B (0,1), F (1) =1$$
Proof $f ' (1) \geq0$.
The geometrical meaning of this topic is very clear, and cannot be rotated around the $1$, otherwise there is no guarantee that the image set is still within the unit circle. Here is a proof of the parse:
Near $1$ we have $ $f (z) =1+f ' (1) (z-1) +o (z-1) $$
Note $|f (z) |<1$, thereby \begin{align*}f (z) \overline{f (z)}&=\left (1+f ' (1) (z-1) +o (z-1) \right) \left (1+\overline{f ' (1) (Z-1)} +\overline{o (z-1)}\right) \\&=1+2{\rm Re}\left (f ' (1) (z-1) \right) +o (|z-1|) <1\\ \rightarrow {\rm re}\left (f ' (1) e^{i\theta}\right) +o (1) &<0\tag{1}\end{align*}
Where $\theta={\rm arg} (z-1) $, apparently $z$ falls on $1$ in $b (0,1) $ in the full small neighborhood $$\theta\in (-\pi,-\frac{\pi}{2}) \cup (\frac{\pi}{2},\pi]$$
In (1) Make $\theta=\pi$ $${\rm re}f ' (1) \geq 0$$
Let $\theta\to\pm\frac{\pi}{2}$ $${\rm re}if ' (1) \leq0,-{\rm re}if ' (1) \leq0$$
The combination of three formulas apparently $f ' (1) \geq0$.
4. Set $f\in H (b (0,1)) $, if present $z_{0}\in B (0,1) \setminus\{0\}$ make $f (z_{0}) \neq0,f ' (z_{0}) \neq0$ and $$|f (z_{0}) |=\max\limits _{|z|\leq |z_{0}|}| F (z) |$$
So $$\frac{z_{0}f ' (z_{0})}{f (z_{0})}>0.$$
Proof Consider the function $ $g (z) =\frac{f (zz_{0})}{f (z_{0})},z\in B (0,1) \cup\{1\}$$ then use the above conclusion.
Two exercises of full-pure function derivative