If a= +i , find the value of 1+a 1-a . - Vidyadip

Question

If $\text{a}=\cos\theta+\text{i}\sin\theta,$ find the value of $\frac{1+\text{a}}{1-\text{a}}.$

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Answer

$\text{a}=\cos\theta+\text{i}\sin\theta, \ \frac{1+\text{a}}{1-\text{a}}$ $=\frac{1+\cos\theta+\text{i}\sin\theta}{1-\cos\theta-\text{i}\sin\theta}$ $=\frac{(1+\cos\theta+\text{i}\sin\theta)(1+\cos\theta+\text{i}\sin\theta)}{(1-\cos\theta-\text{i}\sin\theta)(1+\cos\theta+\text{i}\sin\theta)}$ [Rationalizing the denominator] $=\frac{(1+\cos\theta+\text{i}\sin\theta)(1-\cos\theta+\text{i}\sin\theta)}{(1-\cos\theta)^2-(\text{i}\sin\theta)^2}$ $=\frac{(1+\text{i}\sin\theta)^2-\cos^2\theta}{1-2\cos\theta+\cos^2\theta+\sin^2\theta}$ $=\frac{1+2\text{i}\sin\theta-\sin^2\theta-\cos^2\theta}{1-2\cos\theta+\cos^2\theta+\sin^2\theta}$ $=\frac{1+2\text{i}\sin\theta-1}{1-2\cos\theta+\cos^2\theta+\sin^2\theta} \ \big[\therefore \cos^2\theta+\sin^2\theta=1\big]$ $=\frac{2\text{i}\sin\theta}{1-2\cos\theta+1}$ $=\frac{2\text{i}\sin\theta}{2-2\cos\theta}$ $=\frac{\text{i}\sin\theta}{1-\cos\theta}$ $=\frac{\text{i}2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}}$ $=\frac{\text{i}\cos\frac{\theta}{2}}{\sin^2\frac{\theta}{2}}=\text{i}\cot\frac{\theta}{2}$

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