Show that $\frac{1+z}{1-z}$ = $i\cot(\frac{\theta}{2})$

Question

Show that $\frac{1+z}{1-z}$ = $i\cot(\frac{\theta}{2}), z=cis(\theta)$

This is the next part of the question I posted just before.

I decided to multiply the bottom by its conjugate.

• $\frac{(1+z)(1+b)}{(1-z)(1+b)}$, $b = \frac{|z|}{z}$
• $\frac{1+z+b+|z|}{1-z+b-|z|}$
• $\frac{2+2x}{2-2iy} = \frac{2+2\cos(\theta)}{2-2i\sin(\theta)}$

Where do I go from here?

Answer 1

$$\begin{align*} 1 + z = &=1 + \cos(x) + i \sin(x)\\ & = 2\cos^2\left(\frac x2\right) + i 2 \sin\left(\frac x2\right) \cos\left(\frac x2\right)\\ &=2\cos\left(\frac x2\right)\times\left[\cos\left(\frac x2\right) + i\sin\left(\frac x2\right)\right]\\ & = \color{blue}{2\cos\left(\frac x2\right)\times e^{i\frac x2}} \end{align*}$$

$$\begin{align*} 1 - z = &=1 - \cos(x) - i \sin(x)\\ & = 2\sin^2\left(\frac x2\right) - i 2 \sin\left(\frac x2\right) \cos\left(\frac x2\right)\\ &=2\sin\left(\frac x2\right)\times\left[\sin\left(\frac x2\right) - i\cos\left(\frac x2\right)\right]\\ & = 2\sin \left(\frac x2\right) \left[\cos\left(\frac x2\right) + i\sin \left(\frac x2\right)\right] \times \left(-i\right) \\ &= \color{blue}{2\sin \left(\frac x2\right)e^{i\frac x2} \times (\frac 1i)} \end{align*}$$

$$\frac {1+z}{1-z} = \frac {2\cos\left(\frac x2\right)\times e^{i\frac x2}}{2\sin \left(\frac x2\right)e^{i\frac x2} \times (\frac 1i)} = i \cot \left(\frac x2\right)$$

Answer 2

You made a mistake in the denominator.You should get $\frac {2+2\cos \theta} {-2i\sin \theta}$. Now use the fact that $1+\cos \, \theta=2\sin^{2} (\theta/2)$ and $sin \, \theta =2\sin (\theta /2) \cos (\theta /2)$. You get $\frac {4cos^{2}\theta/2} {-4i \sin (\theta/2)\cos (\theta/2)}=i\cot (\theta/2)$.

Answer 3

This is a nice problem for illustrating the connection between imaginary exponentials and the circular functions. We are given that $ \ z \ = \ cis \ \theta \ = \ e^{ \ i · \theta} \ \ , \ $ so we can manipulate the ratio $$ \frac{1 \ + \ z}{1 \ - \ z} \ \ = \ \ \frac{1 \ + \ e^{ \ i · \theta}}{1 \ - \ e^{ \ i · \theta}} \ · \ \frac{ e^{ \ -i \ · \ \theta /2}}{ e^{ \ -i \ · \ \theta /2}} \ \ = \ \ \frac{e^{ \ -i \ · \ \theta /2} \ + \ e^{ \ i \ · \ \theta /2}}{e^{ \ -i \ · \ \theta /2} \ - \ e^{ \ i \ · \ \theta /2}} \ \ . $$

In terms of imaginary exponentials, the sine and cosine functions are $$ 2i·\sin \theta \ \ = \ \ e^{ \ i \ · \ \theta } \ - \ e^{ \ -i \ · \ \theta } \ \ \ , \ \ \ 2·\cos \theta \ \ = \ \ e^{ \ i \ · \ \theta} \ + \ e^{ \ -i \ · \ \theta } \ \ , $$

from which it follows that $$ \frac{1 \ + \ z}{1 \ - \ z} \ \ = \ \ \frac{2·\cos (\theta / 2)}{-2i· \sin (\theta / 2)} \ \ = \ \ \frac{1}{-i} · \cot( \theta / 2) \ \ = \ \ i · \cot( \theta / 2) \ \ . $$

If you are instead familiar with the hyperbolic sine and cosine functions, $$ 2 ·\sinh \theta \ \ = \ \ e^{ \ \theta } \ - \ e^{ \ - \theta } \ \ \ , \ \ \ 2·\cosh \theta \ \ = \ \ e^{ \ \theta} \ + \ e^{ \ - \theta } \ \ , $$

you will obtain

$$ \frac{1 \ + \ z}{1 \ - \ z} \ \ = \ \ \frac{2·\cosh ( \ i·\theta / 2 \ )}{-2· \sinh ( \ i·\theta / 2 \ )} $$

and will need the hyperbolic-circular identities $ \ \sinh (\theta) \ = \ -i·\sin(i·\theta) \ \ \ , \ \ \ \cosh (\theta) \ = \ \cos(i·\theta) \ $ to complete the derivation: $$ \frac{1 \ + \ z}{1 \ - \ z} \ \ = \ \ \frac{2·\cos ( \ i^2 · \theta / 2 \ )}{-2· [ \ -i·\sin ( \ i^2 · \theta / 2 \ ) \ ]} \ \ = \ \ \frac{\cos ( \ - \theta / 2 \ )}{i·\sin ( \ - \theta / 2 \ ) } \ \ = \ \ \frac{\cos ( \theta / 2 )}{-i·\sin ( \theta / 2 ) } \ \ , $$

leading to the same conclusion as before.