# an analytic function from unit disk to unit disk with two fixed point

prove that if $f:\mathbb{D}\rightarrow\mathbb{D}$ is analytic with two distinct fixed point then $f$ is identity.

i thought if one of the fixed points were zero by schwarz lemma this statement is easily proved.

but what can i do if fixed points were nonzero?

pls don't answer:ur question is duplicated .i've seen other question & answers but i'm still confused.

help pls

thnx

Let $T$ be an automorphism of $\mathbb{D}$. What do you know about $T^{-1}\circ f \circ T$? Can you find a condition on $T$ that reduces the problem to the known case?
• it is a function from $\mathbb{D}\rightarrow\mathbb{D}$ that fixes zero! am i right? – user115608 May 19 '14 at 17:56
• Whether it fixes zero depends on $T$. What condition on $T$ makes it fix zero? – Daniel Fischer May 19 '14 at 18:01
• no sorry!it is a function from $\mathbb{D}\rightarrow\mathbb{D}$ that fixes $T^{-1}$ (fixed point of f)! am i right? – user115608 May 19 '14 at 18:04
• i want $T(0)=$ fixed point of f. – user115608 May 19 '14 at 18:08
• @Twink $f$ has (at least) two fixed points, hence so has $h$. Let $\zeta\neq 0$ be another fixed point. Then you can deduce $c =\:?$ – Daniel Fischer Nov 3 '14 at 11:02