To solve a problem on an awkward region, map the region onto a simple one, usually the unit disk or the upper half-plane. A small toolkit covers many cases: powers open or close sectors, the exponential unrolls strips, and Möbius maps trade half-planes for disks.
Key idea
• $w = \dfrac{z - i}{z + i}$ maps the upper half-plane onto the unit disk $|w| < 1$ (with $i \mapsto 0$); its inverse is $z = i\,\dfrac{1 + w}{1 - w}$.
• $w = z^n$ maps the sector $0 < \arg z < \frac{\pi}{n}$ onto the upper half-plane.
• $w = e^{\pi z/a}$ maps the horizontal strip $0 < \operatorname{Im} z < a$ onto the upper half-plane.
Chain them: map onto an intermediate region first, then onto the target.
• $w = z^n$ maps the sector $0 < \arg z < \frac{\pi}{n}$ onto the upper half-plane.
• $w = e^{\pi z/a}$ maps the horizontal strip $0 < \operatorname{Im} z < a$ onto the upper half-plane.
Chain them: map onto an intermediate region first, then onto the target.
Why it works
For the first map, $|w| = \frac{|z - i|}{|z + i|}$ compares the distance from $z$ to $i$ with the distance to $-i$. A point in the upper half-plane is closer to $i$, so $|w| < 1$; on the real axis the distances are equal, so the boundary goes to the unit circle. For $z^n$, angles multiply by $n$, so the sector's opening $\frac{\pi}{n}$ becomes $\pi$. For the strip, $e^{x + iy}$ sends the line at height $y$ to the ray at angle $y$; heights $0$ to $\pi$ give the upper half-plane, and rescaling by $\frac{\pi}{a}$ fits a strip of height $a$ to that.
Worked examples
Worked example: A point of the half-plane
Where does $w = \dfrac{z - i}{z + i}$ send $z = 1 + 2i$?
- $\frac{1 + i}{1 + 3i} = \frac{(1 + i)(1 - 3i)}{10} = \frac{4 - 2i}{10} = \frac{2}{5} - \frac{1}{5}i$.
- Check: $|w|^2 = \frac{4 + 1}{25} = \frac{1}{5} < 1$, inside the disk as it should be.
Answer: $\frac{2}{5} - \frac{1}{5}i$
Worked example: A strip
Find $k > 0$ so that $e^{kz}$ maps the strip $0 < \operatorname{Im} z < 2$ onto the upper half-plane.
- $e^{\zeta}$ handles a strip of height $\pi$, so stretch heights by $k = \frac{\pi}{2}$.
- $w = e^{\pi z/2}$.
Answer: $k = \frac{\pi}{2}$
Worked example: A chain of maps
Map the first quadrant onto the unit disk, and find the image of $z = 1 + i$.
- $\zeta = z^2$ opens the quadrant ($0 < \arg z < \frac{\pi}{2}$) into the upper half-plane.
- $w = \frac{\zeta - i}{\zeta + i}$ then maps it onto the disk. Together: $w = \frac{z^2 - i}{z^2 + i}$.
- $(1 + i)^2 = 2i$, so $w = \frac{2i - i}{2i + i} = \frac{1}{3}$.
Answer: $w = \frac{z^2 - i}{z^2 + i}$; $\frac{1}{3}$
In applications the order is: map the awkward region onto the upper half-plane or the disk, solve the problem there (where the answer is often known in closed form), and map the solution back. Because the maps are conformal, harmonic functions stay harmonic along the way.
Common mistake
Checking the boundary but not the inside. $\frac{z - i}{z + i}$ and $\frac{z + i}{z - i}$ both send the real axis onto the unit circle, but the first sends the upper half-plane inside the disk and the second sends it outside. After matching boundaries, test one interior point: here $i \mapsto 0$, inside the disk, so the first map is the right one.