# how to show this map by some branch cuts maps unit circle into strip

• Feb 9th 2009, 07:02 PM
silversand
how to show this map by some branch cuts maps unit circle into strip
$\displaystyle f(z)=log(\frac{1+z}{1-z})$ defined by branch cuts on the real axis from negative infinity to -1 and from 1 to infinity, f(0)=0

show by setting $\displaystyle z=e^{i\theta}$ that f maps the disc |z|<1 onto the infinite strip $\displaystyle \{ x+iy : -\frac{\pi}{2}<y<\frac{\pi}{2} \}$
• Feb 9th 2009, 07:13 PM
ThePerfectHacker
Quote:

Originally Posted by silversand
$\displaystyle f(z)=log(\frac{1+z}{1-z})$ defined by branch cuts on the real axis from negative infinity to -1 and from 1 to infinity, f(0)=0

show by setting $\displaystyle z=e^{i\theta}$ that f maps the disc |z|<1 onto the infinite strip $\displaystyle \{ x+iy : -\frac{\pi}{2}<y<\frac{\pi}{2} \}$

You are studying conformal mapping. Go here for a longer discussion. Now in that thread it is show that $\displaystyle z\mapsto \tfrac{1-z}{1+z}$ maps the disk $\displaystyle |z|<1$ onto the right half-plane $\displaystyle \Re (z) > 0$. But you asking about $\displaystyle z\mapsto \tfrac{1+z}{1-z}$. This is not a problem if we do the necessary adjusments. We can think of $\displaystyle z\mapsto \tfrac{1+z}{1-z}$ as first $\displaystyle z\mapsto -z$ and then followed by $\displaystyle z\mapsto \tfrac{1-z}{1+z}$. However, $\displaystyle z\mapsto -z$ is an automorphism of the disk $\displaystyle |z|<1$ and so $\displaystyle z\mapsto \tfrac{1+z}{1-z}$ maps the disk onto the half-plane $\displaystyle \Re (z)>0$. From here it ought to be clear that the log maps the half-plane to a infinite strip (that is what logs always do as conformal mappings).