Question

If a-ib/a+ib=1+i/1-i,then prove that a+b=0. ANSWER ONLY IF YOU KNOW!(RELATED TO COMPLEX NUMBERS) PLS HELP..​

Answer 1

Step-by-step explanation:

$\frac{a - ib}{a + ib} = \frac{1 + i}{1 - i} \\ \frac{a - ib}{a + ib} \times \frac{a - ib}{a - ib} = \frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i} \\(Rationalising\: the \:denominators\: of \\\:both \:sides) \\ \frac{(a - ib)^{2} }{(a) ^{2} - (ib)^{2} } = \frac{(1 + i) ^{2} }{(1)^{2} - (i)^{2}} \\ \frac{ {a}^{2} + (i)^{2}(b)^{2} - 2 \times a \times ib}{ {a}^{2} - {i}^{2} {b}^{2} } = \frac{1 + {i}^{2} + 2i}{1 - ( - 1)}\\... (\because i^2 = - 1)\\ \frac{ {a}^{2} - {b}^{2} - 2abi}{ {a}^{2} + {b}^{2} } = \frac{1 - 1 + 2i}{1 + 1} \\ \frac{ {a}^{2} - {b}^{2} - 2abi }{ {a}^{2} + b^{2} } = \frac{2i}{2} \\ \frac{ {a}^{2} - {b}^{2} - 2abi }{ {a}^{2} + b^{2} } = i \\ \therefore \: {a}^{2} - {b}^{2} - 2abi = i( {a}^{2} + {b}^{2}) \\\therefore \: {a}^{2} - {b}^{2} - 2abi =0 + i( {a}^{2} + {b}^{2}) \\ Equating \: real \: and \: imaginary \: \\ parts \: from \: both \: sides : \\ {a}^{2} - {b}^{2} = 0 \\ {a}^{2} + {b}^{2} = - 2ab...(1) \\ \because \: {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab....(2) \\ \therefore \: from \: equations \: (1) \: and \: (2) \\ {(a + b)}^{2} = - 2ab + 2ab\\ \implies \: {(a + b)}^{2} = 0 \\ \implies \: {a + b} = 0 \\ Thus \: Proved.$