A doubt about an element of $\text{Aut}(\Bbb{H})$

Question

If $\theta \in \Bbb{R}$, this mapping $\varphi(z)=e^{i\theta}z$ is an automorphism of the disc $\Bbb{D}$.And $F(z)=\dfrac{i-z}{i+z}$ is a conformal map from $\Bbb{H}$ to $\Bbb{D}$, its inverse is $F^{-1}(z)=i\dfrac{1-z}{1+z}$.Then $$F^{-1}\circ\varphi\circ F=i\dfrac{(1+e^{i\theta})z+i(1-e^{i\theta})}{(1-e^{i\theta})z+(1+e^{i\theta})}$$ is an automorphism of the upper half plane $\Bbb{H}$. But there is a theorem say: Every automorphism of $\Bbb{H}$ takes the form $\dfrac{az+b}{cz+d}$,where $a,b,c,d \in \Bbb{R}$ and $ad-bc>0$. Apparently,for some $\theta$, the mapping $F^{-1}\circ\varphi\circ F$ can't satisfy this condition. but it is an automorphism of $\Bbb{H}$. I can't figure out the problem. Please help me. thanks very much.

Answer 1

$$i\frac{(1+e^{i\theta})z+i(1-e^{i\theta})}{(1-e^{i\theta})z+i(1+e^{i\theta})}= \frac{~~~\displaystyle\left(\frac{e^{i\theta/2}+e^{-i\theta/2}}{2}\right)z+\left(\frac{e^{i\theta/2}-e^{-i\theta/2}}{2i}\right)}{\displaystyle -\left(\frac{e^{i\theta/2}-e^{-i\theta/2}}{2i}\right)z+\left(\frac{e^{i\theta/2}+e^{-i\theta/2}}{2}\right)}=\frac{~\,~\cos(\theta)z+\sin(\theta)}{-\sin(\theta)z+\cos(\theta)}$$