Find a linear fractional transformation \(T\) which maps the open upper half plane onto the open unit disk. Then explicitly describe the image of the first quadrant of the unit disk under \(T\).
Solution
Let \[ \mathcal H=\{z\in\CC:\operatorname{Im}z>0\}, \qquad \DD=\{w\in\CC:\abs w<1\}, \] and define \[ T(z)=\frac{z-i}{z+i}. \]
1. The map\[ T:\mathcal H\longrightarrow\DD \]is a conformal bijection, with inverse\[ T^{-1}(w)=i\frac{1+w}{1-w}. \]
Write \[ z=x+iy, \qquad y>0. \] Then \[ \abs{z+i}^2-\abs{z-i}^2 = \bigl(x^2+(y+1)^2\bigr) - \bigl(x^2+(y-1)^2\bigr) = 4y > 0. \] Hence \[ \abs{T(z)}<1. \]
Solving \[ w=\frac{z-i}{z+i} \] for \(z\) gives the displayed inverse. The same calculation in reverse shows that this inverse maps \(\DD\) into \(\mathcal H\), so \(T\) is bijective. Finally, \[ T'(z)=\frac{2i}{(z+i)^2}\neq0 \] on \(\mathcal H\). Thus \(T\) is conformal.
2. Let\[ Q=\{z\in\CC:\abs z<1,\ \operatorname{Re}z>0,\ \operatorname{Im}z>0\}, \]the open first-quadrant sector of the unit disk. Then\[ T(Q)\subseteq \{w\in\CC:\abs w<1,\ \operatorname{Re}w<0,\ \operatorname{Im}w<0\}. \]
For \[ z=x+iy\in Q, \] put \[ \Delta=x^2+(y+1)^2>0. \] Multiplying numerator and denominator of \(T(z)\) by \[ x-i(y+1) \] gives \[ \operatorname{Re}T(z) = \frac{x^2+y^2-1}{\Delta} < 0 \] because \(\abs z<1\), and \[ \operatorname{Im}T(z) = \frac{-2x}{\Delta} < 0 \] because \(x>0\). Since \(y>0\), step 1 also gives \(\abs{T(z)}<1\).
3. Conversely, if\[ w\in\DD, \qquad \operatorname{Re}w<0, \qquad \operatorname{Im}w<0, \]then\[ T^{-1}(w)\in Q. \]
Write \[ w=u+iv \] and set \[ z=T^{-1}(w)=i\frac{1+w}{1-w}. \] Put \[ \Gamma=(1-u)^2+v^2>0. \] Direct calculation gives \[ \operatorname{Re}z = \frac{-2v}{\Gamma} > 0 \] because \(v<0\), and \[ \operatorname{Im}z = \frac{1-\abs w^2}{\Gamma} > 0 \] because \(\abs w<1\).
Moreover, \[ \abs z^2 = \frac{\abs{1+w}^2}{\abs{1-w}^2}. \] Since \[ \abs{1-w}^2-\abs{1+w}^2=-4u>0, \] we have \(\abs z<1\). Hence \(z\in Q\).
4. Therefore the image of the open first-quadrant sector of the unit diskis\[ \boxed{ T(Q) = \{w\in\CC:\abs w<1,\ \operatorname{Re}w<0,\ \operatorname{Im}w<0\}, } \]the open third-quadrant sector of the unit disk.
Step 2 gives one inclusion, and step 3 gives the reverse inclusion.
5. The three boundary pieces are sent as follows:\[ \begin{aligned} [0,1]&\longmapsto \text{the unit-circle arc from }-1\text{ to }-i,\\ [0,i]&\longmapsto[-1,0],\\ \{z:\abs z=1,\ \operatorname{Re}z,\operatorname{Im}z\geq0\} &\longmapsto[-i,0]. \end{aligned} \]
The endpoint values are \[ T(0)=-1, \qquad T(1)=-i, \qquad T(i)=0. \] The real axis maps to the unit circle because it is the boundary of \(\mathcal H\); for \(z=iy\) with \(0\leq y\leq1\), \[ T(iy)=\frac{y-1}{y+1}\in[-1,0]; \] and the circle \(\abs z=1\), which passes through the pole \(-i\), maps under a Möbius transformation to a line. Since the relevant arc has endpoint images \(-i\) and \(0\), it maps to the segment \([-i,0]\).
6. Q.E.D.
Step 1 supplies the required linear fractional transformation, and step 4 explicitly describes the requested image; step 5 records its boundary correspondence.