Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can anybody help me to compute the integral

$$\int_0^{2\pi} \frac{1}{z-\cos(\phi)} d\phi$$

where $z \in \mathbb{C}$ denotes a complex number? Thank you!!

share|cite|improve this question

HINT: This is doable by Weierstrass substitution, with an example similar to yours worked out on the linked page.

share|cite|improve this answer
I don't think the worked-out example applies to the situation here, $z$ being a complex number. What to make of the substitution $t = \tan(\phi/z)$? – Minh May 18 '12 at 18:15
The substitution is still $t = \tan(\phi/2)$, and the fact that $z$ is complex is immaterial, as long as $z$ is on on the real interval $[-1,1]$. – Sasha May 18 '12 at 18:30

I suppose that $z \notin [-1,1]$. $$ I = \int_0^{2 \pi} \frac{d \theta}{z - \cos \theta} = \int_0^{2 \pi} \frac{ 2 e^{i \theta} d \theta}{2z e^{i \theta} - (e^{i \theta})^2 - 1} = 2i \int_\gamma \frac{d \zeta}{\zeta^2 - 2 z \zeta + 1} $$ where $\gamma$ is the unitary circle. Since $z \neq \pm 1$, the meromorphic function $f(\zeta) = (\zeta^2 - 2 z \zeta + 1)^{-1}$ has two simple poles $$ \zeta_{1,2} = z \pm \sqrt{z^2 - 1}. $$ Now you have to compute the residues of $f$ in $\zeta_{1,2}$, to check when $\zeta_{1,2}$ are contained in the unitary disk, and to use residue theorem to calculate $I$.

share|cite|improve this answer

This won't finish the problem off, but it's far too long for a "comment", so I'm putting it here.

Initially it seems heartening that the $z$ just stays put as $\varphi$ goes from $0$ to $2\pi$, but later we may have problems.

So I tried the Weierstrass substitution: $$ \sin\varphi = \frac{2t}{1+t^2},\qquad \cos\varphi= \frac{1-t^2}{1+t^2},\qquad d\varphi= \frac{2\,dt}{1+t^2} $$

Then $$ \int_0^{2\pi} \frac{d\varphi}{z-\cos\varphi} = \int_{-\infty}^\infty \frac{2\,dt}{(z+1)t^2 + (z-1)} $$

If you don't see where the bounds of integration come from then think about how the Weierstrass substitution works.

This becomes $$ \frac{2}{z-1}\int_{-\infty}^\infty \frac{dt}{\left(\frac{z+1}{z-1}\right) t^2 + 1} = \frac{2}{\sqrt{z^2-1}}\int_{???} \frac{du}{u^2 + 1} $$

Now we have to worry about the multiple-valued nature of the square-root function and maybe the arctangent function. The path followed by $u$ through the plane will be an unbounded straight line.

The denominator has zeroes at $\pm i$. I think one would look at whether the line along which $u$ moves goes between those or they're both on the same side of it.

Maybe I'll add more here later some day.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.