4-1.Complex numbers
easy

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

A

$2\ i$

B

$1 -i$

C

$-i/2$

D

$i/2$

Solution

(c) Given $z = 1 + i$ and $i = \sqrt { – 1} .$ Squaring both sides, we get ${z^2} = {(1 + i)^2} = 1 + 2i + {i^2} = 1 + 2i – 1$ or ${z^2} = 2i.$
Since it is multiplicative identity, therefore multiplicative inverse of ${z^2} = \frac{1}{{2i}} \times \frac{i}{i} = \frac{i}{{2{i^2}}} = – \frac{i}{2}.$

Standard 11
Mathematics

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