# How can one show that a möbius transformation maps the unit circle to some axis?

Suppose I have the möbius transformation f(z) = $\frac{1+z}{1-z}$ and I want to show that it will map the unit circle (excluding the point 1) to the imaginary axis. Would it help to express my unit circle in polar form?

Here's what I did: f(e$^{i\theta}$) = $\frac{1+e^{i\theta}}{1-e^{i\theta}}$

But I'm not sure where to go from this step.

Point $z=1$ corresponds to $\theta=0$. $e^{i\theta}$ is non zero, and so is $e^{i\theta/2}$. Dividing both numerator and denominator yield: $$f(e^{i\theta})=\frac{e^{-i\theta/2}+e^{i\theta/2}}{e^{-i\theta/2}-e^{i\theta/2}}=\frac{\cos(-\theta/2)+i\sin(-\theta/2)+cos(\theta/2)+i\sin(\theta/2)}{\cos(-\theta/2)+i\sin(-\theta/2)-cos(\theta/2)-i\sin(\theta/2)}=\\ \frac{2\cos(\theta/2)}{-2i\sin(\theta/2)}=-i\cot(\theta/2)$$ For the circle $\theta$ will vary between 0 and 2$\pi$, so $\theta/2$ will vary between 0 and $\pi$, and $\cot(\theta/2)$ will vary between $+\infty$ and $-\infty$. Therefore $f(z)$ will map to all points on the imaginary axis