# Vertical Line through z=0 in complex plane mapped with f(z)=(1+z)/(1-z)

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?

• Please use LaTeX typesetting in your questions. – avs Apr 8 at 22:02
• $f(i)=e^{\pi/4}$ is not correct – J. W. Tanner Apr 8 at 22:05

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: $$-1$$).