Question for finding bound for $f'(z)$[CSIR-December 2011] 



My attempt:-
(1)Taking $f(z)=.5$, So, $g(z)= \begin{cases} 
      \frac{.5}{z} & z\neq 0 \\
     0 & z=0
   \end{cases}
$
So, I can eliminate (1) and (2)
I am trying to apply Schwarz pick lemma for (c), But I am not able to make $|f'(z)|\leq \frac{1-|f(z)|^2}{1-|z|^2}\leq 1$ 
Please help me.
 A: Take $f(z)=z^2$ for which $|f'(3/4)|=3/2\ge 1$. So (c) is false.
Option (d) is true by using Schwarz Pick theorem, 
$|f'(0)|\leq \frac{1-|f(0)|^2}{1-|0|^2}=1-|f(0)|^2\leq 1$ 
A: (3) is false by taking $f(z)=z^2.$
(4) follows from Cauchy's integral theorem: $$f'(0)= \frac{1}{2\pi i} \int_C \frac{f(z)}{z^2} dz,$$ where $C$ is a circle of radius $r < 1$.  Bound $f'(0)$ using the bound for $f$.  Then take $r \to 1^-.$
A: Let $$f(z)=f(0)+zf'(0)+\frac{z^2}{2!}f''(0)+ . . .$$
$(1)\qquad$ Since $f(z)$ is analytic / holomorphic, so from the above equation we have $$f(0)=0, \qquad g(z)= \frac{f(z)}{z}=f'(0)+\frac{z}{2!}f''(0)+ . . .$$
Therefore $g(z)$ is holomorphic on $D$. $\qquad$(Option $1$ is true)
$(2)\qquad$ $$|g(z)|=|\frac{f(z)}{z}|\le 1 \qquad \forall z \in D$$(given that $|f(z)|\le 1$ and $|z|\lt 1$).$\qquad$(Option $2$ is true)
$(3)\qquad$ By Schwarz Pick Lemma, $$|f'(z)|\le \frac{1-|f(z)|^2}{1-|z|^2}$$
Given $$|f(z)|\le 1 \qquad \text{and} \qquad |z|\lt 1$$
So $|f'(z)|$ may or may not be less than or equal to $1\quad \forall z\in D.$$\qquad$(Option $3$ is not  true)
$(4)\qquad$ $$|g(0)|=|f'(0)|\le 1$$ $\qquad$(Option $4$ is true)
