I want to do the following integral in my complex analysis class:

$$ \int_{0}^{2\pi}\frac{d\theta}{3 + \sin\theta + \cos\theta}$$

I don't have the solution (in the textbook) but I checked via WolframAlpha, and they have a solution of $\frac{2\pi}{\sqrt 7}$ but I seem to be getting $\frac{2 \pi (1+i)}{\sqrt 7}$. I won't lay out everything I am doing since that would take me a long time, but basically I set $z = e^{i\theta}$, do all the substitutions, and at some point get: $$ \int_{|z| = 1} \frac{2 \, dz}{6zi + z^2 - 1 + iz^2 + i}$$ I solve for the roots of the denominator, get: $$z_1 = \frac{(1+i)(-3 + \sqrt 7)}{2}, z_2 = \frac{(1+i)(-3 - \sqrt 7)}{2}$$

I realize that $z_1$ falls in the circle I am integrating over, while $z_2$ extends pass the same circle. Then I have the following: $$\int_{|z| = 1}\frac{2 \, dz}{(z-z_1)(z-z_2)}$$ I use Cauchy's formula to integrate and get the solution: $$\frac{4 \pi i}{z_1 - z_2}$$ Then doing the algebra I get the answer I said in the beginning, I am thinking I am doing something wrong with my algebra or maybe there is something more systematically wrong with my method, anyway I want to know how to do this question. Any help would be appreciated!

  • 1
    $\begingroup$ For one thing, your denominator does not factor as $(z-z_1)(z-z_2)$, since the leading coefficient is $(1+i)$ rather than $1$. That explains the extra factor of $(1+i)$. As for the $\sqrt{7}$... $\endgroup$
    – Braindead
    Mar 18 '14 at 2:35
  • $\begingroup$ Oh I am sorry I get $\frac{2 \pi (1+i)}{\sqrt 7}$ $\endgroup$
    – InsigMath
    Mar 18 '14 at 2:36
  • $\begingroup$ Darn I keep making dump little typos, yeah there should be a $\pi$ $\endgroup$
    – InsigMath
    Mar 18 '14 at 2:39
  • $\begingroup$ Are you okay now? Or do you need some further clarification? $\endgroup$
    – Braindead
    Mar 18 '14 at 2:40
  • $\begingroup$ Umm not really... I get what you mean, but then wouldn't my equation then be $(z - (1+i)b)(z - (1+i)c)$? How would the (1+i) factor out so nicely? $\endgroup$
    – InsigMath
    Mar 18 '14 at 2:42

If you have a second degree polynomial with a leading coefficient $a$, then it factors as


The reason is:

  1. The expression has zeros at $z_1$ and $z_2$.
  2. The leading term is going to be $az^2$. Without the $a$, it would be just $z^2$. This is something you should be able to see by looking at the product of the highest degree terms in each factor.

If you are still feeling skeptical, simply multiply out:

$$(1+i)(z - \dfrac{(1+i)(-3+\sqrt{7})}{2})(z - \dfrac{(1+i)(-3-\sqrt{7})}{2})$$

And see what you get.

  • $\begingroup$ I get my original complex polynomial, at least that's what I get when I let wolfram alpha do the dirty work :P ... thanks that helps a lot, this would solve my problem! $\endgroup$
    – InsigMath
    Mar 18 '14 at 3:04
  • $\begingroup$ Basically, if you have $(z-z_1)(z-z_2)$, you would get $z^2 + \dots $. So if your original polynomial is $az^2 + \dots$, you need to have $a(z-z_1)(z-z_2)$. $\endgroup$
    – Braindead
    Mar 18 '14 at 15:21

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