# Automorphism of the Unit Disc Satisfying Properties is the identity

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.

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 $$2 \pi$$].