$f:\mathbb D\rightarrow \mathbb D$ be holomorphic with $f(0)=0$ and $f(1/2)=0.$ I was trying to solve the following problem:
Let $f:\mathbb D\rightarrow \mathbb D$ be holomorphic with $f(0)=0$ and $f(1/2)=0,$ where $\mathbb D=\{z:|z|<1\}.$ Then which of the following statements are correct?

(a)$|f'(1/2)|\leq 4/3,$
(b)$|f'(0)|\leq 1,$
(c)$|f'(1/2)|\leq 4/3$ and $|f'(0)|\leq 1,$
(d)$f(z)=z$ for $z\in \mathbb D.$

My Attempts: Clearly, here we can apply  Schwarz Lemma.By this lemma,we can say that $|f(z)|\leq |z|$ for all $z \in\mathbb D$ and $|f'(0)|\leq 1$ and so i can say option $(b)$ is definitely correct.But i can not say anything about option$(a)$.Please help.Thanks in advance for your time.
 A: First, as @Conan Wong commented, use Schwarz lemma to say that $f(z)=z\cdot g(z)$ 
where $g:\mathbb{D}\to\mathbb{D}$ is holomorphic. Moreover, $g(1/2)=0$. Consider the 
map $\varphi_a=\frac{z-a}{1-\bar{a}z}$. This map is injective, satisfies 
$\varphi_a(\mathbb{D})=\mathbb{D}$ for any $a\in \mathbb{D}$ and $\varphi_a(a)=0$. 
Hence the map $\varphi(z)=\varphi_{1/2}(z)=\frac{2z-1}{2-z}$ preserves $\mathbb{D}$ and sends $1/2$ to $0$.
Look at $h(z)=g(\varphi^{-1}(z))$. We have $h(0)=0$ and $h:\mathbb{D}\to\mathbb{D}$ is holomorphic, 
so we can apply Schwarz lemma to $h$ once again, to get $h(z)=g(\varphi^{-1}(z))=zk(z)$ where $k:\mathbb{D}\to\mathbb{D}$ is holomorphic.
So $g(z)=\varphi(z)k(\varphi(z))$. In conclusion, $f(z)=z\cdot\frac{2z-1}{2-z}\cdot K(z)$ where $K:\mathbb{D}\to\mathbb{D}$ is holomorphic. 
Can you continue?
A: Statement (b) follows from the Schwarz inequality.  A similar trick can be applied for $z = \tfrac{1}{2}$:
$$g(z) = f(z)\frac{2-z}{2z-1}$$ is holomorphic on $\mathbb{D}$ and $|g(z)| \leq 1$.  Also $g(\tfrac{1}{2}) = \tfrac{3}{4}f'(\tfrac{1}{2})$.
