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 :)

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- Jan 18th 2010, 09:53 AMCharlieengineer84Help - 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 :)

- Jan 18th 2010, 10:23 AMJhevon
One way: let $\displaystyle z = x + iy$ where $\displaystyle x,y \in \mathbb{R}$, then we have

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

this implies

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

$\displaystyle 2xy = 2$

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

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

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

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

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

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