Question

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

Answer 1

\[\frac{1 + a}{1 - a} = \frac{1 + \cos\theta + i\sin\theta}{1 - \cos\theta - i\sin\theta}\]

\[ = \frac{(1 + cos\theta) + isin\theta}{(1 - cos\theta) - isin\theta} \times \frac{(1 - cos\theta) + isin\theta}{(1 - cos\theta) + isin\theta}\]

\[ = \frac{1 - \cos\theta + i\sin\theta + \cos\theta - \cos^2 \theta + i\cos\theta\sin\theta + i\sin\theta - i\sin\theta\cos\theta + i^2 \sin^2 \theta}{(1 - \cos\theta )^2 - i^2 \sin^2 \theta}\]

\[ = \frac{1 - \cos^2 \theta - \sin^2 \theta + 2i\sin\theta}{1 + \cos^2 \theta - 2i\cos\theta + \sin^2 \theta}\phantom{.....}...[ \because i^2 = - 1]\]

\[ = \frac{\sin^2 \theta - \sin^2 \theta + 2i\sin\theta}{2 - 2i\cos\theta} \phantom{.....}...[ \because \cos^2 \theta + \sin^2 \theta = 1]\]

\[ = \frac{i\sin\theta}{1 - \cos\theta}\]

\[ = \frac{\cancel{2}i\cancel{\sin\frac{\theta}{2}}\cos\frac{\theta}{2}}{\cancel{2}\cancel{\sin^2 \frac{\theta}{2}}}\]

\[ = i\cot\frac{\theta}{2}\]

Thus, \[\frac{1 + a}{1 - a} = i\cot\frac{\theta}{2}\].