Question

S. T (1+i)^2/1-i - (1-i) ^2/1+i is purely image no.​

Answer 1

Answer:

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Step-by-step explanation:

$to \: show = \\ \frac{(1 + i) {}^{2} }{1 - i} \: + \: \frac{(1 - i) {}^{2} }{1 + i} \: \: is \: purely \: imaginary \\ \\ so \: here \\ \\ = \frac{(1 + i) {}^{2} }{1 - i} \: + \: \frac{(1 - i) {}^{2} }{1 + i} \\ \\ = \frac{1 + i {}^{2} + 2i}{1 - i} \: + \frac{1 + i {}^{2} - 2i }{ 1 + i} \\ \\ = \frac{1 + ( - 1) + 2i}{1 - i} \: + \frac{1 + ( - 1) - 2i}{1 + i} \\ \\ = \frac{2i}{1 - i} \: + \frac{ - 2i}{1 + i} \\ \\ = \frac{2i(1 + i) + 2i(1 - i)}{(1 + i)(1 -i) } \\ \\ = \frac{2i + 2i {}^{2} + (2i - 2i {}^{2} ) }{1 - i {}^{2} } \\ \\ = \frac{2i + 2i {}^{2} + 2i - 2i {}^{2} }{1 - ( - 1)} \\ \\ = \frac{4i }{1 + 1} \\ \\ = \frac{4i}{2} \\ \\ = 2i \\ \\ = 2 \sqrt{ - 1} \\ \\ which \: is \: purely \: imaginary \\ \\ \\ hence \: proved$