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Math Help - how to show this map by some branch cuts maps unit circle into strip

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    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}<y<\frac{\pi}{2} \}
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    Quote Originally Posted by silversand View Post
     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}<y<\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).
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