# Ngô Quốc Anh

## August 14, 2009

### An easy way to construct a conformal mapping between upper half plane and the open unit disk

Thelocus $|z + 1|= | z -1|$ is the perpendicular bisector of the line segment joining $-1$ to $1$, that is, the imaginary axis. The set $|z + 1|< | z -1|$ is then the set of points $z$ closer to $-1$ than to $1$, that is, the left half-plane $\Re z <0$. Hence, $\Re z <0$ if and only if

$\displaystyle\frac{|z+1|}{|z-1|}<1$.

The map

$\displaystyle f : z \mapsto \frac{z+1}{z-1}$

maps $\{z : \Re z < 0\}$ onto $\{w : |w|<1\}$, and is conformal as $f' \ne 0$. The inverse map is easily seen to be

$\displaystyle w \mapsto z=\frac{w+1}{w-1}$.

The locus $|z + i|= | z - i|$ is the perpendicular bisector of the line segment joining $-i$ to $i$, that is, the real axis. The set $|z + i|< | z -i|$ is then the set of points $z$ closer to $-i$ than to $i$, that is, the lower half-plane $\Im z <0$. Hence, $\Im z <0$ if and only if

$\displaystyle \frac{|z+i|}{|z-i|}<1$.

The map

$\displaystyle g : z \mapsto \frac{z+i}{z-i}$

maps $\{z : \Im z < 0\}$ onto $\{w : |w|<1\}$, and is conformal as $g' \ne 0$. The inverse map is easily seen to be

$\displaystyle w \mapsto z=\frac{w+1}{w-1}i$.