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Find a holomorphic function which is also injection such that $$\{z\in \mathbb C: \text{Re}z>0, \text{Im}z>0\} \mapsto \{z\in \mathbb C: |z|<1, \text{Im}z>0\}.$$

I think that I need two transformations: $$\{z\in \mathbb C: \text{Re}z>0, \text{Im}z>0\} \mapsto \{z\in \mathbb C: |z|<1, \text{Re}z>0, \text{Im}z>0\} \mapsto\{z\in \mathbb C: |z|<1, \text{Im}z>0\},$$ but I don't know how to designate them.

vitamin d
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john1235
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1 Answers1

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If $f(z) = \frac{z+1}{z-1}$ then by computation $f^{-1} = f$ and you can see that it sends the upper unit semi-disc to the third quadrant of $\mathbb{C}$ by checking on the boundary: $f(-1)=0, f(1)=\infty, f(0) =-1, f(i) = -i$. You can prove that it's injective by a direct computation.

So if $g(z) = -z$ then $g$ maps the first quadrant of the plane to the third, and thus $f^{-1} \circ g = f \circ g = f(-z) = \frac{z-1}{z+1}$ is a conformal automorphism from your first domain to the other.

John Samples
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  • Btw the little Dover book "A Collection of Problems on Complex Analysis" has a big section on conformal automorphisms if you want more problems. There are also a lot of examples on this site; search "conformal automorphism" to find more problems like this. Eventually you learn there's really only a handful of functions you use; everything else is just shifting/stretching pieces around to make it work. – John Samples Mar 03 '21 at 01:49