Ngô Quốc Anh

August 14, 2009

How to find a conformal mapping between the quadrants and the semidisc

In the previous topic I show you by the following map $\displaystyle f : z \mapsto \frac{z+1}{z-1}$

maps $\{z : \Re z < 0\}$ onto $\{w : |w|<1\}$, and is conformal. Therefore the map $\displaystyle g : z \mapsto \frac{z+1}{1-z}$

maps $\{z : \Re z > 0\}$ onto $\{w : |w|<1\}$, and is conformal. What I am going to do is to prove that $\displaystyle \Re z > 0\quad \Leftrightarrow \quad \Re \left( {\frac{{z + 1}} {{1-z}}} \right) > 0$.

To this purpose, we assume $z=x+iy$, i.e., $\Re z = y$. Now by a simple calculation $\displaystyle\frac{{z+1}}{{1-z}}=\frac{{\left({x+1}\right)+iy}}{{\left({1-x}\right)-iy}}=\frac{{\left[{\left({x+1}\right)+iy}\right]\left[{\left({1-x}\right)+iy}\right]}}{{{{\left({1-x}\right)}^{2}}+{y^{2}}}}$

which yields $\displaystyle \Re \left( {\frac{{z + 1}} {{1 - z}}} \right) = \frac{{2y}} {{{{\left( {1 - x} \right)}^2} + {y^2}}}$.

Having this fact we can easily see that under the map $g$ the first and fourth quadrants maps to upper and lower semidisks, respectively.