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

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

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

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

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

show by setting $z=e^{i\theta}$ that f maps the disc |z|<1 onto the infinite strip $\{ x+iy : -\frac{\pi}{2}
You are studying conformal mapping. Go here for a longer discussion. Now in that thread it is show that $z\mapsto \tfrac{1-z}{1+z}$ maps the disk $|z|<1$ onto the right half-plane $\Re (z) > 0$. But you asking about $z\mapsto \tfrac{1+z}{1-z}$. This is not a problem if we do the necessary adjusments. We can think of $z\mapsto \tfrac{1+z}{1-z}$ as first $z\mapsto -z$ and then followed by $z\mapsto \tfrac{1-z}{1+z}$. However, $z\mapsto -z$ is an automorphism of the disk $|z|<1$ and so $z\mapsto \tfrac{1+z}{1-z}$ maps the disk onto the half-plane $\Re (z)>0$. 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).