Question

Express in the form of `a+ib:(1+i)(1+i)^(-1)`

Answer 1

Answer:

your solution is as follows

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Step-by-step explanation:

$given \: complex \: number \: is \\ z = a + ib = [(1 + i)(1 + i)] {}^{ - 1} \\ \\ = \frac{1}{(1 + i)(1 + i)} \\ \\ = \frac{1}{1 + i {}^{2} + 2i } \\ \\ = \frac{1}{1 - 1 + 2i} \\ \\ = \frac{1}{2i} \\ \\ = \frac{1}{2i} \\ \\ = \frac{1}{2i} \times \frac{ - 2i}{ - 2i} \\ \\ = \frac{ - 2i}{ - 4i {}^{2} } \\ \\ = \frac{ - 2i}{ - 4( - 1)} \\ \\ = \frac{ - 2i}{4} \\ \\ = - \frac{i}{2} \\ \\ a + ib = 0 - \frac{i}{2}$