$Z^2 = \frac{2+5i}{-3+7i}$
I times bottom and top by
${-3-7i}$
coming to
$Z^2=\frac{29-29i}{58} $
ultimately breaking down to
$Z^2= 1/2 +1/2i $
$r^2e^(2i\theta)$ ( i dont know how to put 2 i theta as a power)
for R i got =
$ r^2 = \frac{\sqrt 2}{2}$
$\theta = \frac{\pi}{8}, -\frac{7\pi}{8}$
some help please im sure i did something wrong :(
There is a minor mistake in your simplification of $\dfrac{2+5i}{-3+7i}$; it is equal to $$\dfrac12-\dfrac i2=\dfrac1{\sqrt2}\left(\cos\left(\frac{7\pi}4\right)+\sin\left(\frac{7\pi}4\right)i\right).$$Therefore,$$Z=\pm\frac1{\sqrt[4]2}\left(\cos\left(\frac{7\pi}8\right)+\sin\left(\frac{7\pi}8\right)i\right).$$