Question

Find the square root of the given below complex number:
(ii) $-7-24i$

Answer 1

Let $\sqrt{-7-24i}=x+iy$
$\Rightarrow {\sqrt{(-7-24i)}}^2=(x+iy{)}^2$
{squaring both sides}
$\Rightarrow -7-24i=x^2+i^2{y}^2+2ixy$
$\Rightarrow -7-24i=x^2-y^2+2ixy$
Equating the real and imaginary parts
$\therefore x^2-y^2=-7\dots \left(1\right)$
And $2xy=-24$
$\Rightarrow xy=-24$
$\Rightarrow y=\frac{-12}{x}\dots \left(2\right)$
Put value of 𝑦 in equation (1)
$\therefore x^2-(\frac{-12}{x}{)}^2=-7$
$\Rightarrow x^4+7x^2-144=0$
$\Rightarrow x^4+16x^2-9x^2-144=0$
$\Rightarrow (x^2+16)(x^2-9)=0$
$\Rightarrow x^2-9=0$
$\Rightarrow x=\pm 3$
From equation (2)
$y=\frac{-12}{x}$
$Ifx=3\Rightarrow y=-4$
$x=-3\Rightarrow y=4$
$\therefore x+iy=3-4i,-3+4i$
$\Rightarrow x+iy=\pm (3-4i)$
Therefore, square root of $-7-24i$ is equal to $\pm (3-4i)$