Algebraic manipulation of a complex valued function

Question

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...

Answer 1

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 $$\frac{z}{z-1}= 1 -\frac{1}{2}-\frac{i}{2}\frac{\sin t}{1-\cos t} = \frac{1}{2} -\frac{i}{2}\cot \frac{t}{2}$$

Answer 2

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.

Answer 3

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 $\sqrt z=e^{it},z=?$

$$\dfrac1{z-1}=\dfrac{e^{-it}}{e^{it}-e^{-it}}=\dfrac{\cos t-i\sin t}{2i\sin t}=?$$