Answer to Question #125900 in Complex Analysis for Sanjana

Question #125900
There exists an analytic univalent function f that maps the infinite strip {z : 0 < Im z < 1} onto the unit disk.
1
Expert's answer
2020-07-13T18:57:01-0400

"z_1=\\pi z" maps the horizontal strip "\\{z: \\ 0<\\text{Im } z<1\\}" onto the horizontal strip "\\{z_1: \\ 0<\\text{Im } z_1<\\pi\\}" .

The exponential function "z_2=e^{z_1}" maps the horizontal strip "\\{z_1: \\ 0<\\text{Im } z_1<\\pi\\}" to the half plane "\\{z_2: \\ \\text{Im} \\ z_2>0\\}" .

"z_3=e^{-i \\pi\/2}z_2" rotates the complex plane by "-90^\\circ", "\\{z_3:\\ \\text{Re}\\ z_3>0\\}" .

The linear fractional transformation "w=\\frac{1-z_3}{1+z_3}" maps the right half plane "\\{z_3:\\ \\text{Re} \\ z_3>0\\}" onto the unit disk.

So, we have function "f=\\frac{1-z_3}{1+z_3}= \\frac{1-e^{-i \\pi\/2}z_2}{1+ e^{-i \\pi\/2}z_2}=\\frac{1+i z_2}{1-i z_2}= \\frac{1+i e^{z_1}}{1-i e^{z_1}}= \\frac{1+i e^{\\pi z}}{1-i e^{\\pi z}}" .

Answer: "f(z)= \\frac{1+i e^{\\pi z}}{1-i e^{\\pi z}}."


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