Algebraic manipulation of a complex valued function I have the following function: $$f(z)=\frac{z}{z-1}$$
With complex domain and range, I have to show that the unit circle $e^{i\theta}$ is mapped by the function as a line with real part equal to $\frac{1}{2}$. Moreover i have to show:$$f( e^{i\theta})=\frac{1}{2}-\frac{1}{2}i\cot\frac{\theta}{2} $$
I tried substituting $z=e^{i\theta}$ and manipulating the expression but I keep getting stuck in messy trig expressions...
 A: You can calculate it directly as follows:


*

*$z = \cos t + i \sin t$ (I use $t$ instead of $\theta$.)

*$\frac{z}{z-1}= 1 +\frac{1}{z-1}$
$$\frac{1}{z-1} = \frac{1}{\cos t-1 + i \sin t}= \frac{\cos t-1 - i \sin t}{\cos^2 t +1 -2 \cos t + \sin ^2 t}= \frac{\cos t-1 - i \sin t}{2(1-\cos t)}= -\frac{1}{2}-\frac{i}{2}\frac{\sin t}{1-\cos t}$$
$$\frac{\sin t}{1-\cos t} = \frac{2 \sin \frac{t}{2} \cos \frac{t}{2}}{1 - (\cos^2 \frac{t}{2} - \sin^2 \frac{t}{2})} = \frac{\cos \frac{t}{2}}{\sin \frac{t}{2}}= \cot \frac{t}{2}$$
So, all together
$$\frac{z}{z-1}= 1 -\frac{1}{2}-\frac{i}{2}\frac{\sin t}{1-\cos t} = \frac{1}{2} -\frac{i}{2}\cot \frac{t}{2}$$

A: Hint:
Write
$$
f(z)=\frac{z}{z-1}=1+\frac{1}{z-1}=g^{-1}(h(g(z)))
$$
where
$$
g(z)=z-1,
\qquad
h(z)=\frac{1}{z}
$$
Since $g$ and $g^{-1}$ are translations, the only hard part is the inversion of a circle into a line.
A: If $z=\cos2u+i\sin2u,$
$$\dfrac1{z-1}=\dfrac1{\cos2u-1+i\sin2u}=\dfrac1{2i\sin u(\cos u+i\sin u)}=\dfrac{\cos u-i\sin u}{2i\sin u}=?$$
Alternatively if $\sqrt z=e^{it},z=?$
$$\dfrac1{z-1}=\dfrac{e^{-it}}{e^{it}-e^{-it}}=\dfrac{\cos t-i\sin t}{2i\sin t}=?$$
