Finding a Möbius transformation, why do I need to expand area? I'm currently studying for an re-exam in complex analysis, and got a question regarding Möbius transformation.
The exam-question is following:
Find a conform and bijective mapping from $A := ${$ z: 0 < |z| < 1, 0 < Arg(z) < \frac{\pi}{2}$}
to $B := ${$z: |z| < 1, |z-1-i| < 1$}.
I've been able to draw the area and visualise it, but no more than that. I looked into the solution given, and to solve it they first extend the area $A$ with $z^2$ so that $A' := ${$ z: 0 < |z| < 1, Im(z) > 0$} and then mapping $A'$ to $B$.
My question is, what exactly makes us want to extend the area $A$ to be able to solve it? The only reason I can come up with is that we want the same number of curves that limits the area. Is this the reason, or am I missing out on something?
 A: Sources to read for problems like this
I've written a couple of other answers recently about how to find conformal maps between specified regions.

*

*General approaches for finding such conformal maps here

*"Constructing conformal mapping from $\mathbb{C} \setminus \{\text{2 discs plus a line segment}\}$" here

*"Constructing conformal mapping from union of three overlapping discs" here
Note that for (2) and (3) you may need to go into the edit history, because the OP keeps editing their own post to remove the diagram and other useful info?? Anyway, I think these examples cover most of the standard tricks for problems like this.
Specific reasoning for the current problem
The intuition behind the step you're asking about is: You have a region that's bounded by 3 circles. (Lines count as circles.) By squaring, you can turn it into a region bounded by 2 circles instead, which is simpler.
Here's how I'd do this problem, focusing on reasons behind each step and skipping over some of the computation details.

*

*$z mapsto z^2$. Now we have a half-disc, bounded by 2 circles.

*Mobius transformation: if my region is bounded by 2 circles, I instantly want to apply a Mobius transformation that sends 1 of the intersection points to 0 and the other to $\infty$. (You still have 1 extra degree of freedom on that Mobius transformation but it doesn't matter too much.) After this step, we'll have a quarter-plane.

*$z \mapsto z^2$ again. This takes us from a quarter plane (bounded by 2 circles) to a half plane (bounded by 1 circle, as simple as it gets).

To map $B$ to the upper half plane I'd use similar logic.

*

*We're starting with a region bounded by 2 circles. Find the intersection points and use a Mobius transformation to map one intersection point to $0$ and the other to $\infty$. We'll end up with a quarter-plane again.

*Square the quarter-plane to get a half-plane.

Final note: in general, the trick about mapping the region between two circles by sending intersection points to 0 and $\infty$ will not always create a quarter-plane. It will always be a nice region between two rays that start at $0$, but the region could have a different size depending on what the original circles were. For example, if you started with the region $\{z \in \mathbb{C} : |z|<1 \text{ and } |z-1|<1\}$ then the trick would move you to a "sixth of the plane" like $\{z \in \mathbb{C} : \text{arg}(z) \in (0, \frac \pi 3)\}$. From there you'd still want to get a half-plane, but this time you should use $z \mapsto z^3$ instead of $z \mapsto z^2$. Do you see why?
