Let $a,b,c,d\in \mathbb{R}$ be such that $ad-bc>0$. Consider the Möbius transformation $$T_{a,b,c,d}(z)=\frac{az+b}{cz+d}$$ and define $$\begin{align}H_{+}&=\{z\in\mathbb{C}:\Im(z)>0\},\\H_{-}&=\{z\in\mathbb{C}:\Im(z)<0\},\\R_{+}&=\{z\in\mathbb{C}:\Re(z)>0\},\\R_{-}&=\{z\in\mathbb{C}:\Re(z)<0\}. \end{align}$$
Then $T_{a,b,c,d}$ maps
$$\begin{align}(a)&\quad H_{+} \text{ to } H_{+} \\ (b)&\quad H_{+} \text{ to } H_{-} \\ (c)&\quad R_{+} \text{ to } R_{+} \\ (d)&\quad R_{+} \text{ to } R_{-}\end{align}$$
Usually we know that $$z\in H_{+} \rightarrow \frac{1}{z}\in H_{-}$$ but I didn't get any further... How do we go about solving this problem?
We notice that $T_{a,b,c,d}$ maps $\Bbb R\cup\{\infty\}$ to itself. Hence (a) and (b) are possible options. As $T_{a,b,c,d}(i)=\frac{ai+b}{ci+d}=\frac{(ai+b)(-ci+d)}{(ci+d)(-ci+d)}=\frac{(bd+ac)+(\color{red}{ad-bc})i}{c^2+d^2}$ has positive imaginary part, we see that (a) is certainly valid and (b) false. On the other hand, (c) and (d) may or may not be valid.