TITLE: Application of Schwarz lemma and Liouville's theorem
QUESTION [5 upvotes]: I recently came across the following problem

Let $h : \Bbb C → \Bbb C$ be an analytic function such that $h(0) = 0, h(1/2) = 5$, and $|h(z)| < 10$ for $|z| < 1$. Then pick out the correct statement(s):

the set $\{z : |h(z)| = 5\}$ is unbounded by the Maximum Principle
the set $\{z : |h'(z)| = 5\}$ is a circle of strictly positive radius
$h(1) = 10$
regardless of what $h'$ is, $h'' \equiv 0$.


My try
Consider $f(z)=\frac{h(z)}{10}$ and $f$ satisfies the hypothesis of Schwarz lemma, so $$\vert f(z) \vert \leq \vert z \vert$$ and so $$\vert h(z) \vert \leq 10 \vert z \vert ^1$$ so by extended Lioville's theorem, $h$ is a polynomial of degree atmost $1$. So $h$ is either constant or linear. But given hypothesis implies $h$ is not constant and $h$ is exactly $10z$. That is, $$(\forall z \in \Bbb C): h(z)\equiv 10z$$
So the set in the first bullet is a circle $\vert z \vert =1/2$ ,which is bounded and so it is false
The second set is $\varnothing$ and so the second  one is false
Third and fourth are true. Since $h$ is linear , its second derivative vanishes!
Is my reasoning correct ? Any thoughts?
Edit:  I agree that my reasoning in wrong as Mr. Kavi Rama Murthy says in the comment. So this $h(z)=10z$ helps to eliminate the first two options. But how to conclude the last two. Any help ?

REPLY [2 votes]: Let $g(z)= h(z)/10.$ Then $g$ restricted to $\mathbb D,$ let's call it $\tilde g,$ maps $\mathbb D$ into $\mathbb D,$ sends $0$ to $0,$ and takes $1/2$ to $1/2.$ By the Schwarz Lemma, $\tilde g(z)= z.$ That is the same as saying $g(z)=z$ in $\mathbb D.$ By the identity principle, $g(z) = z$ on all of $\mathbb C.$ Therefore $h(z) = 10z$ on $\mathbb C.$ In order then, we see the given statements are false, false, true, and true.