The map \[ f\colon\ts{z\st\abs{z}<1,\ \Im z>0}\to\ts{w\st\Re w>0,\ \Im w>0},\qquad f(z)=\frac{1+z}{1-z}, \] is a biholomorphism from the upper half-disc onto the first quadrant, with inverse \(f^{-1}(w)=\frac{w-1}{w+1}\).
For \(z=x+iy\) with \(x^2+y^2<1\) and \(y>0\), multiplying numerator and denominator by \(1-\bar z\) gives \[ f(z)=\frac{(1+z)(1-\bar z)}{\abs{1-z}^2}=\frac{1-(x^2+y^2)}{(1-x)^2+y^2}+i\,\frac{2y}{(1-x)^2+y^2}, \] and both parts are positive, so \(f(z)\) lies in the first quadrant.
For \(w=u+iv\) with \(u,v>0\), the point \(w\) is closer to \(1\) than to \(-1\) because \(u>0\), so \(\abs{\frac{w-1}{w+1}}<1\). Moreover \[ \frac{w-1}{w+1}=\frac{(u-1+iv)(u+1-iv)}{(u+1)^2+v^2}=\frac{u^2+v^2-1}{(u+1)^2+v^2}+i\,\frac{2v}{(u+1)^2+v^2}, \] whose imaginary part is positive. So \(w\mapsto\frac{w-1}{w+1}\) maps the first quadrant into the upper half-disc, and solving \(w=\frac{1+z}{1-z}\) for \(z\) shows that the two maps are mutually inverse.
On the upper unit semicircle, for \(z=e^{i\theta}\) with \(0<\theta<\pi\), \[ f(e^{i\theta})=\frac{e^{-i\theta/2}+e^{i\theta/2}}{e^{-i\theta/2}-e^{i\theta/2}}=\frac{2\cos(\theta/2)}{-2i\sin(\theta/2)}=i\cot(\theta/2), \] so as \(\theta\) increases from \(0\) to \(\pi\), \(f(e^{i\theta})\) moves from \(i\infty\) to \(0\) along the positive imaginary axis. As \(x\) increases from \(-1\) to \(1\), \(f(x)=\frac{1+x}{1-x}\) increases from \(0\) to \(\infty\) along the positive real axis, with \(f(0)=1\).