# Thread: Help - Square rooting a complex number

1. ## Help - Square rooting a complex number

z^2 = 2√3 + 2j

I just need pushing in the right direction?? I am sure its simpler than I think it is!!

Thank you

2. Originally Posted by Charlieengineer84
z^2 = 2√3 + 2j

I just need pushing in the right direction?? I am sure its simpler than I think it is!!

Thank you
One way: let $z = x + iy$ where $x,y \in \mathbb{R}$, then we have

$(x + iy)^2 = 2 \sqrt 3 + 2i$

this implies

$x^2 - y^2 = 2 \sqrt 3$ and,

$2xy = 2$

Solve this system simultaneously for $x$ and $y$ and you will find $z$.

Another way: use polar coordinates (or the cis form if you prefer)

Let $z = re^{i \theta} \implies z^2 = r^2 e^{2 i \theta}$.

Thus, $z^2 = 2 \sqrt 3 + 2 i$

$\Rightarrow r^2 e^{2i \theta} = 4 \left( \frac {\sqrt 3}2 + \frac 12i \right) = 4e^{i(\pi / 6 + 2k \pi)}$

Now, solve for $r$ and $\theta$ ( $k = 0, 1$)