This question was asked in my complex analysis quiz and I was unable to solve it due to lack of time. I am looking for an alternative method than mine.
Question: Let $a$, $b$, $c$, $d\in\mathbb{C}$ be such that $ad-bc>0$. Consider the Möbius transformation $T(z)= \frac{az+b} {cz+d}$ . Define $H_{+}=\{z \in \mathbb{C}\mid Im(z)>0\}$, $H_{-}=\{z \in\mathbb{C}\mid Im(z)<0\}$, $R_{+}=\{z \in\mathbb{C}\mid Re(z)>0\}$, $R_{-}=\{z \in\mathbb{C}\mid Re(z)<0\}$:
Then $T(z)$ maps
(i) $H_{+} \mapsto H_{+}$
(ii)$ H_{+} \mapsto H_{-}$
(iii) $R_{+} \mapsto R_{+}$
(iv)$R_{+} \mapsto R_{-}$
Its a single correct answer.
Attempt : I tried by writing $z$ as $x+iy$ in $T(z)$ and then removing iota from the denominator by multiplying and dividing by $(cx+d-iy)$ but I think thats a very lenghty method.
Can you please tell me alternative and relatively short method.
Thanks!!