= Solution
The domains are <conformally equivalent>. The <Möbius transformation>
$$
T(z)=\frac{1+z}{1-z}
$$
maps the upper half of the unit disc onto the first quadrant: its diameter $(-1,1)$ maps to the positive real axis and its upper semicircle maps to the positive imaginary axis. Squaring maps the first quadrant bijectively and conformally onto the upper half-plane. Therefore
$$
\boxed{F(z)=\left(\frac{1+z}{1-z}\right)^2}
$$
is the required equivalence, as recorded by the <conformal equivalence between the upper half-disc and upper half-plane>.
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