How can I prove that $A_1$ is conformally equivalent to $A_2$?

[Euge]

Here's this week's problem!

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Problem. Let $A_1 = \{z \in \Bbb C : |z| < 1\}$ and $A_2 = \{z \in A_1 : \operatorname{Im}(z) > 0\}$. Prove $A_1$ is conformally equivalent to $A_2$.

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[Jameson]

Euge has asked me to fill in this week. :)
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No one answered this week's problem correctly. Here is the solution from Euge:

Spoiler

The function $S(z) = \frac{1 + z}{1 - z}$ maps $A_2$ conformally onto the first quadrant. Squaring will map the first quadrant conformally onto the upper-half plane. Composing with $T(z) = \frac{z - i}{1 - iz}$ will map the upper-half plane conformally onto $A_1$. So the function $F(z) = T(S(z)^2)$ is a conformal equivalence of $A_2$ with $A_1$.