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I need to show that the only automorphism of the unit disc with $f(0)=0$, $f'(0)>0$ is the identity map. If f satisfies those properties then, $f(z)=e^{i\theta}z$, since the automorphisms of unit disc are of the form $$e^{i\theta} (\frac{z-\alpha}{1-\overline{\alpha}z})\,,\:|\alpha|<1.$$ Then $f'(z)=e^{i\theta}=f'(0)>0$. But I'm not sure where to go from here.

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When is $e^{i\theta }>0$ (for $\theta $ real)? What points on the unit circle are positive numbers? Answer: $e^{i\theta}$ must be $1$ so $f(z)=z$.

[You may also know that $\cos \theta=1$ and $\sin \theta =0$ are possible only when $\theta $ is an integer multiple of $2 \pi$].

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