Question

Find the square root of complex number $-4-3i$.

• A. $\pm \frac{1}{\sqrt{2}}(3+i)$
• B. $\pm \frac{1}{\sqrt{3}}(3+i)$
• C. $\pm \frac{1}{\sqrt{3}}(3+2i)$
• D. $\pm \frac{1}{\sqrt{3}}(4+i)$

Answer 1

The correct option is A $\pm \frac{1}{\sqrt{2}}(3+i)$

Let $\sqrt{-4-3i}=a+ib$

$-4-3i=a^2-b^2+2aib$

$a^2-b^2=-4,ab=\frac{-3}{2}$

$ab=\frac{-3}{2}\Rightarrow b=\frac{-3}{2a}$

$a^2-b^2=-4\Rightarrow a^2-\frac{9}{4a^2}=-4$

$4a^4-9=-16a^2$

$4a^4+16a^2-9=0$

$4a^4+18a^2-2a^2-9=0$

$2a^2(2a^2+9)-1(2a^2+9)=0$

$a=\pm \frac{1}{\sqrt{2}},a=\pm i\frac{3}{\sqrt{2}}$

$b=\pm \frac{3}{\sqrt{2}},b=\pm \frac{1}{2}$

Therefore, $\sqrt{-4-3i}=\pm \frac{1}{\sqrt{2}}(1+3i)$

or $\pm \frac{1}{\sqrt{2}}(3+i)$