Question

If (1+i)^3/(1-i)^3 - (1-i)^3/(1+i)^3 = x+iy , find x and y
class 11 complex no.s

Answer 1

Answer:

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Step-by-step explanation:

$so \: here \\ \frac{(1 + i) {}^{3} }{(1 - i) {}^{3} } \: - \: \frac{(1 - i) {}^{3} }{(1 + i) {}^{3} } = x + iy \\ \\ ( \frac{1 + i}{1 - i} ) {}^{3} \: - \: ( \frac{1 - i}{1 + i} ) {}^{3} = x + iy \\ \\ so \: here \: ( \frac{1 + i}{1 - i} ) {}^{3} \: - \: ( \frac{1 - i}{1 + i} ) {}^{3} \: \: is \: in \: form \: a {}^{3} - b {}^{3} \\ so \: we \: know \: that \\ a {}^{3} - b {}^{3} = (a - b)(a {}^{2} + b {}^{2} + ab) \\ \\ thus \: accordingly \\ \\ = ( \: \frac{1 + i}{1 - i} \: - \: \frac{1 - i}{1 + i} \: )(( \frac{1 + i}{1 - i} ) {}^{2} + ( \frac{1 - i}{ 1+i } \: ) {}^{2} + ( \: \frac{1 + i}{1 - i} \times \frac{1 - i}{1 + i} \: )) \\ \\ = ( \frac{(1 + i)(1 + i) - (1 - i)(1 - i)}{(1 + i)(1 - i)} )( \frac{1 + i {}^{2} + 2i}{1 + i {}^{2} - 2i } \: + \: \frac{1 + i {}^{2} - 2i}{1 + i {}^{2} + 2i} \: + \: 1) \\ \\ = ( \frac{1 + i {}^{2} + 2i - (1 + i { }^{2} - 2i)}{1 - i {}^{2} } )( \frac{1 + ( - 1) +2 i}{1 + ( - 1) - 2i} \: + \: \frac{1 + ( - 1) - 2i}{1 + ( - 1) + 2i \: } \: + \: 1 \: ) \\ \\ = ( \frac{1 + i {}^{2} + 2i - 1 - i {}^{2} +2 i}{1 - ( - 1)} )( \frac{1 - 1 + 2i}{1 - 1 - 2i} \: + \: \frac{1 - 1 - 2 i}{1 - 1 + 2i} \: + \: 1) \\ \\ = ( \frac{2i + 2i}{1 +1 } )( \frac{2i}{ - 2i} \: + \frac{( - 2i)}{2i} \: + \: 1 ) \\ \\ = ( \frac{4i}{2} )( - 1 + ( - 1) + 1) \\ \\ = 2i( - 1) \\ = - 2i \\ = - 2i + 0 \\ \\ comparing \: the \: imaginary \: part \: and \: real \: part \\ \: we \: get \\ x = 0 \: \: \: y = - 2$