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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.

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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

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The beauty of linear fractional transformations (Mobius transformations) is that they map symmetric points to (again) symmetric points.

So, if you can you show that a pair of pre-image points symmetric w.r.t. the unit circle, map to a pair of points symmetric w.r.t. the imaginary axis, then you are done.

Always look for symmetry arguments.

If you want an excellent reference, read the section written by Ahlfors.

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