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,

  • $\begingroup$ I'd like to add a follow-up question: since these are my four zeroes, and we know that every polynomial over the complex numbers splits, and has at most four zeroes (not necessarily distinct), then I can write the polynomial as g(z)=(z-z1)(z-z2)(z-z3)(z-z4), where the zi's are my roots. then 1/g(z), which is my integrand in the contour integration problem, has four simple poles. Does this sound ok? Basically, I'm trying to determine the nature of the singularities, in order to proceed with my computation of the residues of f(z). Thanks, $\endgroup$ – User001 Nov 16 '14 at 0:23

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. $\theta = \pi \pm \arccos(\frac{3}{4})$

  • $\begingroup$ Ok, got it. Thanks so much, @CheungSW. Just one comment: I assume you meant to say "equate it with (-3 +/- isqrt(7)) / 2." $\endgroup$ – User001 Nov 16 '14 at 21:18
  • $\begingroup$ Yep. Thanks for your reminder. I planned to work out the + case first but then forgot to update them accordingly. $\endgroup$ – Empiricist Nov 17 '14 at 3:15

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