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 $\displaystyle z = x + iy$ 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 $\displaystyle \theta$ ($\displaystyle k = 0, 1$)