Question

If $x+iy=\frac{a+ib}{a-ib}$ prove that x² + y² = 1.

Answer 1

Solution:

$x + iy = \frac{a + ib}{a - ib}$
rationalised the denominator

$x + iy = \frac{a + ib}{a - ib} \times \frac{a + ib}{a - ib} \\ \\ = \frac{ {(a + ib)}^{2} }{ {a}^{2} + {b}^{2} } \\ \\ = \frac{ {a}^{2} - {b}^{2} + 2iab}{ {a}^{2} + {b}^{2} } \\ \\ x + iy = \frac{ {a}^{2} - {b}^{2} }{{a}^{2} + {b}^{2}} + \frac{2iab}{{a}^{2} + {b}^{2}}$
compare terms of LHS and RHS

$x = \frac{{a}^{2} - {b}^{2}}{{a}^{2} + {b}^{2}} \\ \\ y = \frac{2ab}{{a}^{2} + {b}^{2} } \\ \\ for \: {x}^{2} + {y}^{2} = \Big({\frac{{a}^{2} - {b}^{2}}{{a}^{2} + {b}^{2}}}\Big)^{2} + \Big({\frac{2ab}{{a}^{2} + {b}^{2} }}\Big)^{2} \\ \\ = \frac{ {a}^{4} + {b}^{4} - 2 {a}^{2} {b}^{2} + 4 {a}^{2} {b}^{2} }{ {( {a}^{2} + {b}^{2} })^{2} } \\ \\ = \frac{ {a}^{4} + {b}^{4} + 2 {a}^{2} {b}^{2}}{ {( {a}^{2} + {b}^{2} })^{2} } \\ \\ = \frac{ {( {a}^{2} + {b}^{2}) }^{2} }{{( {a}^{2} + {b}^{2}) }^{2}} \\ \\ = 1 \\ \\ so \\ \\ {x}^{2} + {y}^{2} = 1$

hence proved