If $a + i b = \frac{c + i}{c - i}$ , where $c$ is a real number, then prove that : $a^{2} + b^{2} = 1$ and $\frac{b}{a} = \frac{2 c}{c^{2} - 1}$ .

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Solution

Given,
$a + i b = \frac{c + i}{c - i}$
$= \frac{c + i}{c - i} \times \frac{c + i}{c + i}$ (On rationalisation)
$= \frac{c^{2} - 1 + 2 c i}{c^{2} + 1}$
$= (\frac{c^{2} - 1}{c^{2} + 1}) + (\frac{2 c}{c^{2} + 1}) i$
On comparison,
$a = \frac{c^{2} - 1}{c^{2} + 1}$ and $b = \frac{2 c}{c^{2} + 1}$
Therefore,
$a^{2} + b^{2} = \frac{(c^{2} - 1)^{2} + (2 c)^{2}}{(c^{2} + 1)^{2}}$
$= \frac{c^{4} - 2 c^{2} + 1 + 4 c^{2}}{(c^{2} + 1)^{2}}$
$= \frac{(c^{2} + 1)^{2}}{(c^{2} + 1)^{2}}$ = $1$
Now,
$\frac{b}{a} = \frac{\frac{2 c}{c^{2} + 1}}{\frac{c^{2} - 1}{c^{2} + 1}}$
$\frac{b}{a} = \frac{2 c}{c^{2} - 1}$

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