A Möbius transformation maps the points $0$, $-i$, and $\infty$ to $10$, $5-5i$, and $5+5i$, respectively. The question is to find the image of the set $S = \{ z: \operatorname{Re}(z) < 0 \}$.
I have tried doing this using the cross ratio method and that yields a longish calculation. The trouble is this is from a previous competitive examination and is supposed to be done usually under $3$ minutes. Is there any other way to do it?
The set of lines and circles is preserved by any Mobius transformation. The unique line $L$ passing through $0$, $-i$ and $\infty$ must therefore be mapped to the unique circle $C$ passing through $10$, $5-5i$, $5+5i$, and that circle $C$ is the circle of radius $5$ centered on the point $5$. So the set $S$, which is one of the two half-planes bounded by $L$, must be mapped to either the inside of $C$ or the outside of $C$.
You could therefore complete this problem by picking a single point of $S$ and seeing where it goes: to the inside or to the outside of $C$?
Alternatively, you could also notice that the orientation of $L$ that starts at $0$ and points downward through $-i$ heading towards $\infty$ must be taken to the orientation of $C$ that starts at $10$ and points clockwise through $5-5i$ heading towards $5+5i$. That tells you that $S$, which is the left side of $L$, must map to the inside of $C$.