# Show that if there exist two complex numbers $a,b$ such that $f(a)=a$ and $f(b)=b$ then $f(z)=z$ for all $z\in B(0,1)$.

Let $f:B(0,1) \to B(0,1)$ holomorphic. Show that if there exist two complex numbers $a,b$ such that $f(a)=a$ and $f(b)=b$ then $f(z)=z$ for all $z\in B(0,1)$.

There is a suggestion in the excercise that says, consider the function

$g(z)=\frac{h(z)-a}{1-\overline{a}h(z)}$ with $h(z)=f \left(\frac{z+a}{1+\overline{a}z} \right)$

and use Schwarz Lemma.

Ok, so I've been thinking about this excercise for a while, I wasn't able to solve it. Im not seeing how can I use the suggestion. By replacing, I easily got that $g(0)=0$ but Im not being able to prove that $|g(z)|<1$ and also Im not being able to continue even assuming that is true. Any hint in how to use the suggestion?

Given $c$ with $|c| < 1$, you can easily prove that the map $$h_c(z) = \frac{z - c}{1 - \overline{c}z}$$ is a holomorphic automorphism of the open unit disc sending $c$ to $0$. The inverse of $h_c(z)$ is $h_{-c}$.
If $a$ or $b$ were $0$, the answer would follow from an immediate application of the Schwarz lemma. The hint offers you to reduce the problem to that case by using a holomorphic automorphism of $B(0,1)$ that sends $a$ to $0$. More specifically, you are offered to consider $g = h_a \circ f \circ h_{-a} = h_c \circ f \circ (h_a)^{-1}$ - the conjugation of $f$ by $h_a$. It follows that since $f$ maps the open unit disc into itself, that $|g(z)| < 1$ for all $z \in B(0,1)$. We also have $$g(0) = h_a(f(h_{-a}(0))) = h_a(f(a)) = h_a(a) = 0$$ and $$g(h_a(b)) = h_a(f(h_a^{-1}(h_a(b))) = h_a(f(b)) = h_a(b).$$
Since $h_a(b) \neq 0$ (because $h_a$ is an automorphism and $h_a(a) = 0$), the Schwarz lemma implies that $g(z) = z$ for all $z \in B(0,1)$. The only map conjugate to the identify is the identity itself and so $f(z) = z$ for all $z \in B(0,1)$.
Proving $|g(z)|<1$ is proved by proving the equivalent statement $|g(z)|^2=g\bar{g}<1$. If you write this inequality out and simplify a bit, you get $$|h(z)|^2+|a|^2<1+|a|^2|h(z)|^2.$$ We know this is true since $a$, Im$f\subseteq B(0,1)$, hence implies $|g(z)|<1$. More specifically, multiply over $|h|^2<\frac{1-|a|^2}{1-|a|^2}=1$ and you will get the above inequality.