Question

Find the square root of $2+2\sqrt{3}i$.

Answer 1

le $(2+2\sqrt{3}i{)}^{1/2}=x+iy$

sq. both side

$2+2\sqrt{3}i=x^2-y^2+2xyi$

$\Rightarrow x^2-y^2=2----\left(i\right)$ & $xy=\sqrt{3}----\left(ii\right)$

we know

$(x^2+y^2{)}^2=(x^2-y^2{)}^2+(2xy{)}^2$

$=2^2+(2\sqrt{3}{)}^2$

$=16$

$x^2+y^2=4----\left(iii\right)$

Solve $\left(i\right)$ & $\left(iii\right)$ from $\left(2\right)$ & $\left(iii\right)$

$2x^2=16$ $2xy=2\sqrt{3}$

$x=\pm \sqrt{3}$ $x=\pm \sqrt{3}$

$\therefore \Rightarrow y=\pm 1$

Then,

$(2+2\sqrt{3}i{)}^{1/2}=\pm (\sqrt{3}+i)$