Question

Find the square root of $\frac{-2-14i}{1+i}$

Answer 1

Given,

$\frac{-2-14i}{1+i}$

$=\frac{-2-14i}{1+i}\times \frac{1-i}{1-i}$

$=\frac{-16-12i}{2}$

$=-8-6i$

$(a+ib)(a+ib)=-8-6i$

$(a^2-b^2)+i\left(2ab\right)=-8-6i$

$a^2-b^2=-8$

$2ab=-6\Rightarrow ab=-3$

$\therefore a=-\frac{3}{b}$

${(-\frac{3}{b})}^2-b^2=-8$

$9-b^4=-8b^2$

$b^4-8b^2-9=0$

since b is real, we get, $b=3,-3$

$a^2-(-3{)}^2=-8\Rightarrow a=\pm 1$

$\therefore a=\pm 1,b=\mp 3$

$\therefore (-1+3i),(1-3i)$ is the square root of $\frac{-2-14i}{1+i}$