Question

All the points in the set S = {(α + i)/(α - i) : α ∈ R} (i = √(-1) lie on a:
(A) straight line whose slope is 1 (B) circle whose radius is √2
(C) straight line whose slope is –1 (D) circle whose radius is 1

Answer 1

Answer:

D

First simplfy your point so that we can get it in form of 【 x + i y 】

$\frac{a + i}{a - i} \times \frac{a + i}{a + i} \\ \\ = \frac{(a + i) {}^{2} }{ {a}^{2} + 1 } \\ \\ = \frac{ {a}^{2} - 1 }{ {a}^{2} + 1} + \frac{2a}{ {a}^{2} + 1 } i \\ \\ = x + iy \\ \implies\\ x = \frac{ {a}^{2} - 1}{ {a}^{2} + 1 } \: \: and \: \: y = \frac{2a}{ {a}^{2} + 1} \\ \\ to \: \: get \: \: locus \: eliminate \: \: a \\ \\ let \: \: a = \tan( \alpha ) \\ \\ x = \frac{ \tan {}^{2} ( \alpha ) - 1 }{ \tan {}^{2} ( \alpha ) + 1} \\ = \frac{ \tan {}^{2} ( \alpha ) - 1}{ \sec {}^{2} ( \alpha ) } \\ \\ = \sin {}^{2} ( \alpha ) - \cos {}^{2} ( \alpha ) \\ = - \cos(2 \alpha ) \\ \\ and \\ \\ y = \frac{2 \tan( \alpha ) }{ \tan {}^{2} ( \alpha ) + 1} \\ \\ = \frac{2 \tan( \alpha ) }{ \sec {}^{2} ( \alpha ) } = 2 \sin( \alpha ) \cos( \alpha ) \\ \\ = \sin(2 \alpha ) \\ \\ now \: square \: and \: add \: both \: \\ \\ {x}^{2} + {y }^{2} =( - \cos(2 \alpha ) ) {}^{2} + \sin {}^{2} (2 \alpha ) \\ \\ \huge \pink{{x}^{2} + {y}^{2} = 1}$