Question

(1-i)^-1 simplify.... ​

Answer 1

★solution•

$= (1 - i) {}^{ - 1}$

we know that

$= > a {}^{ - m} = \frac{1}{a {}^{m} } \\ \\ = \frac{1}{(1 - i) {}^{1} } \\ \\ = \frac{1}{1 - i} \\ \\ now \: muliplie consider \: and \\ \: numrator \: (1 + i) \\ \\ = \frac{1 + i}{(1 - i)(1 + i)}$

we know that

$= > (a + b)(a - b) = a {}^{2} - b {}^{2}$

$= \frac{1 + i}{1 {}^{2} -( i) {}^{2} } \\ \\ we \: know \: that \\ i {}^{2} = - 1 \\ \\ put \: iota \: value$

$= \frac{1 + i}{1 - ( - 1)} \\ \\ = \frac{1 + i}{1 + 1} \\ \\ = \frac{1 + i}{2} \\ \\ = \frac{1}{2} + \frac{1}{2} i$

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I hope it helps you