Question

Find the square roots of the following complex numbers:
$8-15i$

Answer 1

Let $a+ib=\sqrt{8-15i}$

${(a+ib)}^2=8-15i.$

$a^2-b^2+2abi=8-15i.$

$\therefore a^2-b^2=\mathrm{8.....}\left(1\right)2ab=-15.$

We know, ${(a^2+b^2)}^2={(a^2-b^2)}^2+{\left(2ab\right)}^2=64+225=289.$

$\therefore a^2+b^2=\mathrm{17.....}\left(2\right)$

From$\left(1\right)$ and $\left(2\right)$

$a^2-b^2=8a^2+b^2=17\_\_\_\_\_\_\_\_2a^2=25\Rightarrow a^2=\frac{25}{2}\Rightarrow a=\pm \frac{5}{\sqrt{2}}$

And $b=\pm \frac{3}{\sqrt{2}}$

$\therefore ab=$ negative value

$\therefore$ aquare root $=\pm \frac{5}{\sqrt{2}}\pm \frac{3}{\sqrt{2}}$