Question

complex number problem
$1 - i \alpha \div 1 + i \alpha = a + ib find the value of {a}^{2} + {b}^{2}$

Answer 1

Ans.

$\frac{1 - i \alpha }{1 + i \alpha } \\ = \frac{(1 - i \alpha )(1 - i \alpha )}{(1 + i \alpha )(1 - i \alpha )} \\ = \frac{1 - 2i \alpha - { \alpha }^{2} }{1 + { \alpha }^{2} } \\ = \frac{1 - { \alpha }^{2} }{1 + { \alpha }^{2} } + i( \frac{ - 2 \alpha }{1 + { \alpha }^{2} } )$
= a + ib

So,

${a}^{2} + {b}^{2} \\ = ( \frac{1 - { \alpha }^{2} }{1 + { \alpha }^{2} } )^{2} +( \frac{ - 2 \alpha }{1 + { \alpha }^{2} } )^{2} \\ = \frac{(1 - 2 { \alpha }^{2} + { \alpha }^{4}) + 4 { \alpha }^{2} }{(1 + { \alpha }^{2} ) ^{2} } \\ = \frac{1 + 2 { { \alpha }^{2} + { \alpha }^{4} } }{ \: 1 + 2 { { \alpha }^{2} + { \alpha }^{4} }} \\ = 1$

I HOPE THAT THIS HELPS YOU.