show that $f$ does not have a zero in the disc $\{z:|z|<|a|\}$. 
Consider the unit disc $D$ and an analytic function $f:D\to D$. If $f(0)=a\not=0$ then show that $f$ does not have a zero in the disc $\{z:|z|<|a|\}$.

My Try:
Consider $g(z)=f(z)-a$. Then $g:D\to D$ is analytic and $g(0)=0$. Now apply Schwarz lemma on $g$ and we get , $|g(z)|<|z|\implies|f(z)-a|<|z|$.
Now , when $|z|<|a|$ then $|f(z)|<2|a|$. From here how we can say that $f$ does not have zero in the given domain ?
Or any other way ?

You are correct to think of the Schwarz lemma.
Define $g(z) = \dfrac{z - a}{1 - \bar a z}$ so that $g : D \to D$ is analytic and $g(a) = 0$. If $h(z) = g(f(z))$ then $h : D \to D$ and $h(0) = 0$ so by the Schwarz lemma $|h(z)| \le |z|$ for all $z \in D$. Thus $$\left| \frac{f(z) - a}{1 - \bar a f(z)} \right| \le |z|$$ for all $z \in D$ and in particular $$|f(z) - a| \le |z| |1 - \bar a f(z)|$$ for all $z \in D$. If $|z| < |a|$ and $f(z) = 0$ this yields $|a| < |a|$.
