Question

If a+ib = c+i / c-i where C is real then prove that a²+b² = 1

Answer 1

$a+ib=\dfrac{c+i}{c-i}$
Taking mod, then squaring both sides,
$|a+ib|^2=\left|\dfrac{c+i}{c-i}\right| ^2\\ a^2+b^2=\dfrac{c^2+1}{c^2+1}\\ =1$

Answer 2

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