Question

Find the square root of the given below complex number:
(i)$5+12i$

Answer 1

Let $\sqrt{-5+12i}=x+iy$
$\Rightarrow ({\sqrt{-5+12i)}}^2=(x+iy{)}^2$
{squaring both sides}
$\Rightarrow -5+12i=x^2+i^2{y}^2+2ixy$
$\Rightarrow -5+12i=x^2-y^2+2ixy$
Equating the real and imaginary parts
$\Rightarrow x^2-y^2=-5\dots \left(1\right)$
And 2𝑥𝑦=12
$\Rightarrow xy=6$
$\Rightarrow y=\frac{6}{x}\dots \left(2\right)$
Put value of 𝑦 in equation (1)
$\therefore x^2-(\frac{6}{x}{)}^2=-5$
$\Rightarrow x^2-\frac{36}{x^2}=-5$
$\Rightarrow x^2-\frac{36}{x^2}=-5$
$\Rightarrow x^4+5x^2-36=0$
$\Rightarrow x^4+9x^2-4x^2-36=0$
$\Rightarrow (x^2+9)(x^2-4)=0$
$\Rightarrow x^2-4=0$
$\Rightarrow x^2=4$
$\Rightarrow x=\pm 2$
From equation (2)
$y=\frac{6}{x}$
If $x=2\Rightarrow y=3$
$x=-2\Rightarrow y=-3$
$\therefore x+iy=2+3i,-2-3i$
$\Rightarrow x+iy=\pm (2+3i)$
Therefore, square root of $-5+12i$ is equal to $\pm (2+3i)$