TITLE: When does a Möbius transformation map $\Im(z)>0$ to itself?
QUESTION [7 upvotes]: Show Möbius transformation which maps $\Im(z)>0$ to itself iff
$$ f(z)= \frac{az+b}{cz+d}\,,\,\,ad-bc>0$$ and $a,b,c,d$ are real.

REPLY [4 votes]: Let $H_{+}=\{z\in \mathbb C: \Im (z)>0\}$ and let $H_{-}=\{z\in \mathbb C: \Im (z)<0\}$. Since $H_{+}$ is connected, therefore, $f(H_{+})$ is connected. Thus $f(H_{+})$ can not intersect both $H_{+}$ and $H_{-}$. Therefore, $f(H_{+})\subset H_{+}$ or $f(H_{+})\subset H_{-}$. 
Since $f(i)=\dfrac{ai+b}{ci+d}=\dfrac{ac+bd+i(ad-bc)}{c^2+d^2} \in H_{+}$. Hence $f$ maps $H_{+}$ onto $H_{+}$