If $z = 1 + i$ , then the multiplicative inverse of $z^{2}$ is (where $i = \sqrt{- 1})$ .

2 i

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1 - i

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- \frac{i}{2}

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\frac{i}{2}

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Solution

The correct option is C $- \frac{i}{2}$
$1$ & $2 = 1 + i$
$z = (1 + i)^{2} = 1 + i^{2} + 21$
$= 0 + 2 i$
As we know multiplicative inverse of $z = a + i b$ is given by
$z^{- 1} = \frac{a}{a^{2} + b^{2}} + i (\frac{- b}{a^{2} + b^{2}})$
multiplicative inverse of $z^{2}$
$= 0 + i (\frac{- 2}{0 + 4})$
$= \frac{- i}{2}$

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