I'm preparing to take an exam in complex analysis, and have come across a particularly (at least, I find) hard past paper example of using contour integration to perform an integral of a function built up from cosines.

The first piece of the question is using Cauchy's residue theorem to find the value of

$\int_\gamma \frac{1}{z^2 - 4z + 3} dz$,

where $\gamma$ is the positively oriented circular contour of radius 2, entered at the origin.

The hard piece of the question is using the above to evaluate the integral

$\int_{\theta=0}^{2\pi} \frac{2 - \cos\theta}{5 - 4\cos\theta} d\theta$.

I begin by writing the cosines in complex exponential form. The question basically boils down to manipulating the $\theta$ integral to look like

$A \int_{\theta=0}^{2\pi} \frac{2ie^{i \theta}}{(2 e^{i \theta})^2 - 4 (2 e^{i \theta}) + 3} d\theta$,

where $A$ is some constant, since I can directly use the first result. However, I just can't see how to manipulate the integral into this form. Can anyone help shed some light on this? Particularly, I am looking for useful thought processes and rules of thumb to help with such manipulations, rather than the answer. Thanks!


Let $I=\int_0^{2\pi} \frac{2-\cos\theta}{5-4\cos\theta}\,d\theta$. Using the substitution $z=2e^{i\theta}$, then $d\theta=\frac{dz}{iz}$. We have $$\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}=\frac{2e^{i\theta}+4\cdot \frac12 e^{-i\theta}}{4}=\frac{z+4/z}{4}.$$

We now have the following contour integral:

$$ \begin{align*}I=\int_{C(0,2)^+} \frac{2-\frac{z+4/z}{4}}{5-4\cdot\frac{z+4/z}{4}}\frac{dz}{iz} &=\int_{C(0,2)^+} \left(\frac14+\frac{3/4}{5-(z+4/z)}\right)\frac{dz}{iz} \\ &= \frac{1}{4i}\int_{C(0,2)^+}\frac1z\,dz+\frac{3}{4i}\int_{C(0,2)^+}\frac{1}{5-z-4/z}\frac1z\,dz \\ &=\frac{1}{4i}\cdot 2\pi i -\frac{3}{4i}\int_{C(0,2)^+}\frac{1}{z^2-5z+4}\,dz. \end{align*} $$

Can you continue?

  • $\begingroup$ Thanks, I can get it from there! Is the step where you manage to write the integrand as $1/4 - (3/4)/(5 - (z + 4/z))$ by polynomial division? $\endgroup$ – M. Whyte Dec 3 '19 at 21:48
  • 1
    $\begingroup$ @M.Whyte yes, exactly! $\endgroup$ – rae306 Dec 3 '19 at 21:49

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