The automorphism of the disc is $\varphi(z)=\exp(i\theta)\frac{\alpha-z}{1-\overline{\alpha}z}$,where $\theta \in \Bbb{R}$ and $\alpha \in \Bbb{D}$ (we denote the open unit disc centered at the origin by $\Bbb{D}$).
Let $F(z)=\frac{i-z}{i+z}$ and $F^{-1}(z)=i\frac{1-z}{1+z}\;$ , then $F:\Bbb{H}=\{z\in\Bbb{C}:\Im(z)>0\}\rightarrow \Bbb{D}$ is a conformal map and $F^{-1}$ is its inverse.
Consider the map
$$\Gamma:\text{Aut}(\Bbb{D})\rightarrow \text{Aut}(\Bbb{H})$$
given by
$$\Gamma(\varphi)=F^{-1}\circ\varphi\circ F$$
How to verify that $\text{Aut}(\Bbb{H})$ consists of all maps $$z \longmapsto \frac{az+b}{cz+d}$$ where $a,b,c,d \in \Bbb{R}$ and $ad-bc=1$?
you can find that the map defines an isomorphism between Aut(D) and Aut(H)(it is easy to prove).later,you just need do some computation,you will find out the answer.specificly,you can find the answer on page 222 in complex analysis written by Stein in Princeton. hope it can help you