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I have the vague notion that the imaginary axis maps to a circle with f(z)=(1+z)/(1-z). $$ \begin{array}{lll} f(\infty) & = & -1\\ f(i) & = & e^{\pi/4}\\ f(-i) & = & e^{-\pi/4}\\ f(0) & = & 1\\ \end{array} $$ Seems to be mapping only in the right side of a unitary circle. Is this correct?

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  • $\begingroup$ Please use LaTeX typesetting in your questions. $\endgroup$ – avs Apr 8 at 22:02
  • $\begingroup$ $f(i)=e^{\pi/4}$ is not correct $\endgroup$ – J. W. Tanner Apr 8 at 22:05
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No. It maps the imaginary axis onto the unit circle minus $-1$. Note that, if $t\in\mathbb R$,$$\frac{1+it}{1-it}=\frac{(1+it)^2}{(1+it)(1-it)}=\frac{1-t^2}{1+t^2}+\frac{2t}{1+t^2}i$$and the complex numbers of this form are precisely the elements of the unit circle (with one exception: $-1$).

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