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.

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