If a = (cosθ + i sinθ), prove that

Solution:

Given: a = cosθ + isinθ

To prove: $\frac{1+a}{1-a}=\left(\cot \frac{\theta}{2}\right) i$

Taking LHS,

$\frac{1+a}{1-a}$

Putting the value of a, we get

$=\frac{1+\cos \theta+i \sin \theta}{1-(\cos \theta+i \sin \theta)}$

$=\frac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}$

We know that,

$1+\cos 2 \theta=2 \cos ^{2} \theta$

Or $1+\cos \theta=2 \cos ^{2} \frac{\theta}{2}$

And $1-\cos \theta=2 \sin ^{2} \frac{\theta}{2}$

Using the above two formulas

$=\frac{2 \cos ^{2} \frac{\theta}{2}+i \sin \theta}{2 \sin ^{2} \frac{\theta}{2}-i \sin \theta}$

Using, $\sin \theta=2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$

$=\frac{2 \cos ^{2} \frac{\theta}{2}+i 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin ^{2} \frac{\theta}{2}-2 i \sin \frac{\theta}{2} \cos \frac{\theta}{2}}$

$=\frac{2 \cos \frac{\theta}{2}\left[\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}\right]}{2 \sin \frac{\theta}{2}\left[\sin \frac{\theta}{2}-i \cos \frac{\theta}{2}\right]}$

$=\cot \frac{\theta}{2}\left[\frac{\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}}{\sin \frac{\theta}{2}-i \cos \frac{\theta}{2}}\right]\left[\because \frac{\cos \theta}{\sin \theta}=\cot \theta\right]$

Rationalizing by multiply and divide by the conjugate of $\sin \frac{\theta}{2}-i \cos \frac{\theta}{2}$

$=\left(\cot \frac{\theta}{2}\right)\left[\frac{\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}}{\sin \frac{\theta}{2}-i \cos \frac{\theta}{2}} \times \frac{\sin \frac{\theta}{2}+i \cos \frac{\theta}{2}}{\sin \frac{\theta}{2}+i \cos \frac{\theta}{2}}\right]$

$=\left(\cot \frac{\theta}{2}\right) \frac{\left(\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}\right)\left(\sin \frac{\theta}{2}+i \cos \frac{\theta}{2}\right)}{\left(\sin \frac{\theta}{2}-i \cos \frac{\theta}{2}\right)\left(\sin \frac{\theta}{2}+i \cos \frac{\theta}{2}\right)}$

$=\left(\cot \frac{\theta}{2}\right) \frac{\left(\cos \frac{\theta}{2}\right)\left(\sin \frac{\theta}{2}+i \cos \frac{\theta}{2}\right)+i \sin \frac{\theta}{2}\left(\sin \frac{\theta}{2}+i \cos \frac{\theta}{2}\right)}{\left(\sin \frac{\theta}{2}\right)^{2}-\left(i \cos \frac{\theta}{2}\right)^{2}}$

$=\left(\cot \frac{\theta}{2}\right) \frac{\cos \frac{\theta}{2} \sin \frac{\theta}{2}+i \cos ^{2} \frac{\theta}{2}+i \sin ^{2} \frac{\theta}{2}+i^{2} \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{\sin ^{2} \frac{\theta}{2}-i^{2} \cos ^{2} \frac{\theta}{2}}$

Putting $i^{2}=-1$, we get

$=\left(\cot \frac{\theta}{2}\right) \frac{\cos \frac{\theta}{2} \sin \frac{\theta}{2}+i \cos ^{2} \frac{\theta}{2}+i \sin ^{2} \frac{\theta}{2}+(-1) \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{\sin ^{2} \frac{\theta}{2}-(-1) \cos ^{2} \frac{\theta}{2}}$

$=\left(\cot \frac{\theta}{2}\right) \frac{\cos \frac{\theta}{2} \sin \frac{\theta}{2}+i\left(\cos ^{2} \frac{\theta}{2}+\sin ^{2} \frac{\theta}{2}\right)-\sin \frac{\theta}{2} \cos \frac{\theta}{2}}{\sin ^{2} \frac{\theta}{2}+\cos ^{2} \frac{\theta}{2}}$

We know that,

$\cos ^{2} \theta+\sin ^{2} \theta=1$

$=\left(\cot \frac{\theta}{2}\right)\left[\frac{i\left(\cos ^{2} \frac{\theta}{2}+\sin ^{2} \frac{\theta}{2}\right)}{1}\right]$

$=\cot \frac{\theta}{2}(i)$

= RHS

Hence Proved