how to show this is a conformal map

• Feb 10th 2009, 09:43 PM
szpengchao
how to show this is a conformal map
$\displaystyle f(z)=\frac{z}{(z-1)^{2}}$ maps disc |z|<1 to $\displaystyle C\ \{x+yi: x\leq -\frac{1}{4}, y=0 \}$
• Feb 11th 2009, 02:48 PM
ThePerfectHacker
Quote:

Originally Posted by szpengchao
$\displaystyle f(z)=\frac{z}{(z-1)^{2}}$ maps disc |z|<1 to $\displaystyle C\ \{x+yi: x\leq -\frac{1}{4}, y=0 \}$

Let $\displaystyle f_1$ be the map $\displaystyle z\mapsto \tfrac{1+z}{1-z}$.
Let $\displaystyle f_2$ be the map $\displaystyle z\mapsto z+1$.
Let $\displaystyle f_3$ be the map $\displaystyle z\mapsto -\tfrac{1}{2}z$.
Let $\displaystyle f_4$ be the map $\displaystyle z\mapsto z^2$.
Let $\displaystyle f_5$ be the map $\displaystyle z\mapsto z+1$.

Then, $\displaystyle z\mapsto \tfrac{z}{(z-1)^2}$ is $\displaystyle f_5\circ f_4\circ f_3 \circ f_2 \circ f_1$.

1)The conformal map $\displaystyle f_1$ maps the disk $\displaystyle |z|<1$ to the half-plane $\displaystyle \Re (z) > 0$.
2)The conformal map $\displaystyle f_2$ maps the half-plane $\displaystyle \Re(z) > 0$ to $\displaystyle \Re(z) > 1$.
3)The conformal map $\displaystyle f_3$ maps the half-plane $\displaystyle \Re(z) > 1$ to $\displaystyle \Re(z) < - \tfrac{1}{2}$.
4)The problem is that $\displaystyle f_4$ is not conformal on this half-plane. (Surprised)