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4-1.Complex numbers
easy
જો $z = 1 + i$ તો $z^2$ નો વ્યસ્ત હોય તો . . . (કે જ્યાં $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 ${z^2} = \frac{1}{{2i}} \times \frac{i}{i} = \frac{i}{{2{i^2}}} = – \frac{i}{2}.$
Std 11
Mathematics
