Find the square root of $\left(-8-6i\right)$
Let $\sqrt{-8 - 6i} = x + iy$ $\;\;\; \cdots \; (1)$
Squaring equation $(1)$ we get,
$-8 - 6i = \left(x^2 - y^2\right) + 2 i xy$ $\;\;\; \cdots \; (2)$
Equating the real parts in equation $(2)$ we have,
$-8 = x^2 - y^2$ $\;\;\; \cdots \; (3a)$
Equating the imaginary parts in equation $(2)$ we have,
$-6 = 2xy$ $\implies$ $xy = -3$ $\;\;\; \cdots \; (3b)$
Now, $\left(x^2 + y^2\right)^2 = \left(x^2 - y^2\right)^2 + 4 x^2 y^2$
i.e. $\;$ $x^2 + y^2 = \sqrt{\left(x^2 - y^2\right)^2 + 4 x^2 y^2}$
$\therefore$ $\;$ We have from equations $(3a)$ and $(3b)$
$x^2 + y^2 = \sqrt{\left(-8\right)^2 + 4 \times \left(-3\right)^2} = \sqrt{100}$
i.e. $\;$ $x^2 + y^2 = 10$ $\;\;\; \cdots \; (4)$
Solving equations $(3a)$ and $(4)$ simultaneously we have,
$2x^2 = 2$ $\implies$ $x^2 = 1$ $\implies$ $x = \pm 1$
Substituting the values of $x^2$ in equation $(4)$ we get,
$y^2 = 9$ $\implies$ $y = \pm 3$
$\because$ $\;$ The product $xy$ is negative [equation $(3b)$]
$\implies$ when $x$ is +ve, $y$ is -ve and when $x$ is -ve, $y$ is +ve.
$\therefore$ $\;$ $\sqrt{-8-6i} = 1-3i$ $\;$ or $\;$ $\sqrt{-8-6i} = -1 + 3i$