I'm trying to locate my four zeroes of a complex-valued function, in order to apply the Residue Theorem.
After using the quadratic formula, I am left with $$z^2 = [-3 \pm i\sqrt7] / 2$$
writing the left side in exponential form, I get: $$z^2 = 2exp(\pm i\theta)/2$$
which gives $$z=\pm \sqrt2exp(\pm i\theta/2)/2$$
these are my four roots; however, I don't know how to compute $\theta$ explicitly. I tried using that arctan(y/x) formula, and but I'm getting anything that's clean.
Thanks in advance,
I do not think there is a $/2$ in the second and the third formulae, as $|z^2|^2 = \frac{(-3)^2 + 7}{2^2} = \frac{16}{4} = 4$, i.e. $|z^2| = 2$ and $z^2$ can be expressed as $z^2 = 2\exp(i\theta)$.
Express $z^2 = 2\exp(i\theta)$ as
$$ z^2 = 2\exp(i\theta) = 2\cos(\theta) + 2i \sin(\theta) $$
and equate it with $\frac{(-3 \pm i\sqrt{7})}{2}$, we obtain
$$ \cos(\theta) = \frac{-3}{4} $$ $$ \sin(\theta) = \pm\frac{\sqrt{7}}{4} $$
i.e. $\theta = \pi \pm \arccos(\frac{3}{4})$