Question:
If $\sqrt{a+i b}=x+i y$, then possible value of $\sqrt{a-i b}$ is
(a) $x^{2}+y^{2}$
(b) $\sqrt{x^{2}+y^{2}}$
(c) $x+i y$
(d) $x-i y$
(e) $\sqrt{x^{2}-y^{2}}$
Solution:
(d) $x-i y$
$\sqrt{a+i b}=x+i y$
Squaring on both the sides, we get,
$a+i b=x^{2}+(i y)^{2}+2 i x y$
$\Rightarrow a+i b=\left(x^{2}-y^{2}\right)+2 i x y$
$\therefore a=\left(x^{2}-y^{2}\right)$
and $b=2 x y$
$\therefore a-i b=\left(x^{2}-y^{2}\right)-2 i x y$
$\Rightarrow a-i b=x^{2}+i^{2} y^{2}-2 i x y \quad\left[\because i^{2}=-1\right]$
Taking square root on both the sides, we get:
$\sqrt{a-i b}=x-i y$
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