Show that any conformal self-map of the upper half-plane has the form

$\displaystyle f(z)=\frac{az+b}{cz+d}$, $\displaystyle \text{Im}(z)>0$,

where $\displaystyle a, b, c, d$ are real numbers satisfying $\displaystyle ad-bc=1$. When do two such coefficient choices for $\displaystyle a, b, c, d$ determine the same conformal self-map of the upper half-plane?

In this section we have covered Pick's Lemma and we have covered that fractional linear transformations have the form $\displaystyle w=f(z)=\frac{az+b}{cz+b}$. However, I do not see how conformal self-maps in the upper half-plane have this form too. I need some hints on how to start this one. Thanks in advance.