Find a fractional linear transformation \(T\) which maps \(\HH\) to \(\DD\), and explicitly describe the image of the first quadrant under \(T\).
Solution
Goal: Find a Möbius transformation \(T: \HH \to \DD\) and describe the image of the first quadrant \(Q = \theset{x + iy \suchthat x > 0, y > 0}\) under \(T\).
1. \(T(z) = \frac{z - i}{z + i}\) is a Möbius transformation mapping \(\HH\) conformally onto \(\DD\).
This is the standard Cayley map: it sends the boundary \(\RR \cup \theset{\infty}\) of \(\HH\) to the unit circle (for real \(z\), \(\abs{z - i} = \abs{z + i}\)), sends \(i \in \HH\) to \(0 \in \DD\), and is a bijection of the Riemann sphere; Möbius maps send circles/lines to circles/lines and preserve the side of the boundary, so \(\HH\) maps onto the component of the complement of the unit circle containing \(0\), i.e. \(\DD\).
2. The first quadrant \(Q\) is the intersection of \(\HH\) with the right half-plane \(\Re z > 0\), bounded by the two rays \(\arg z = 0\) (positive real axis) and \(\arg z = \pi/2\) (positive imaginary axis).
\(Q = \theset{z \suchthat \Re z > 0, \Im z > 0}\).
3. \(T\) maps the positive real axis to the lower semicircle \(\theset{e^{i\theta} \suchthat -\pi < \theta < 0}\) from \(1\) to \(-1\).
For \(x > 0\) real, \(T(x) = \frac{x - i}{x + i}\) has modulus 1 and \(T(1) = \frac{1-i}{1+i} = -i\); as \(x\) runs from \(0^+\) to \(\infty\), \(T(x)\) moves on the unit circle from \(T(0) = -1\) through \(-i\) to \(T(\infty) = 1\), i.e. along the lower semicircle.
4. \(T\) maps the positive imaginary axis to the real interval \((-1, 1)\).
For \(y > 0\), \(T(iy) = \frac{iy - i}{iy + i} = \frac{y-1}{y+1} \in (-1, 1)\), and as \(y\) goes from \(0^+\) to \(\infty\) this runs over all of \((-1, 1)\) (from \(-1\) to \(1\)).
5. \(T(Q)\) is the lower half-disk \(\theset{w \in \DD \suchthat \Im w < 0}\).
\(Q\) is the region between the two rays of 2; its image under the Möbius map \(T\) is the region bounded by the two image curves of 3 and 4, namely the lower semicircle and the diameter \((-1, 1)\).
The point \(i \in Q\) maps to \(T(i) = 0\), which lies in the lower half of the disk (indeed \(0\) is on the boundary between the two halves; take instead a sample point, e.g. \(1 + i \in Q\) with \(T(1+i) = \frac{1}{2+i} = \frac{2 - i}{5}\), which has negative imaginary part), so the image is the lower half-disk.
6. Q.E.D.
1 gives the map \(T(z) = \frac{z-i}{z+i}\) from \(\HH\) onto \(\DD\), and 5 describes the image of the first quadrant as \(\DD \cap \theset{\Im w < 0}\).