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If $\Omega$ be a bounded open subset of $C$, and $\phi :\Omega\to\Omega$ be a holomorphic function satisfied $f(z_0)=z_0,f'(z_0)=1$ then prove that $\phi$ is linear.

The hint told me to let $z_0=0$, and $\phi(z)=z+Cz^n+O(z^{n+1})$, consider $\phi_k=\phi\circ\phi\circ...\circ\phi$ (k times), then prove $\phi_k(z)=z+kCz^n+O(z^{n+1})$, use Cauchy's inequality and let $k\to\infty$ to get $C=0$.

I don't know how to prove $\phi_k(z)=z+kCz^n+O(z^{n+1})$ and here is where i have trouble: assume that in $B(0,r_1)$ we have power series expansion of $\phi$: $\phi(z)=z+Cz^n+O(z^{n+1})$ to make sure we have composite $\phi(\phi)$, we need to restrict $z$ in $D=\{z\in B(0,r_1):\phi(z)\in B(0,r_1)\}$ and assume the maximal disc of $\phi(\phi)$ that has a power series expansion in $D$ is $B(0,r_2)$, then only in this disc we can have $\phi_2=z+2Cz^n+O(z^{n+1})$, next consider $\phi_3$ and so on.

Now we have a series of disc $B_k=B(0,r_k)$ that only in $B_k$ we have the power series expansion $\phi_k(z)=z+kCz^n+O(z^{n+1})$, so what if $r_n$ tend to zero?

We can't let $k\to \infty$.

janmarqz
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Idele
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  • It is probably simpler to apply the Schwarz lemma if you can. – lhf Nov 03 '15 at 02:08
  • @lhf i think to use schwarz lemma we need to find a disc.$D$ contained in $\Omega$ centered at zero and $\phi(D)$ contained in D, how to find such D? – Idele Nov 03 '15 at 05:32
  • by scaling a disc containing $\phi(D)$ ? – lhf Nov 03 '15 at 09:49
  • @lhf I don't see how this could lead to $\phi(D)\subset D$, with the same disk $D$ on both sides. –  Nov 09 '15 at 06:36
  • (Link taken from the Related column on the right). –  Nov 09 '15 at 06:43

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