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I want to evaluate the integral

$$\displaystyle\oint_C\left(\frac{1}{\cos(z)-1}\right)dz$$

where $C$ is circle with radius $7$, counterclockwise.

I kept having problems with it. If someone can help, it would really be appreciated.

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  • $\begingroup$ Your title says the $-1$ should be in the denominator, but the display-line equation in your question body asks something different. Which is it? $\endgroup$
    – J.G.
    Commented Jun 11, 2018 at 19:35
  • $\begingroup$ What is the center of the circle? $\endgroup$
    – user539887
    Commented Jun 11, 2018 at 19:35
  • $\begingroup$ Oh I completely missed the -1 in the denominator, my apologies. I would suppose it should be down there. $\endgroup$
    – mallan
    Commented Jun 11, 2018 at 19:41

2 Answers 2

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We'll make use of residues to avoid using direct integration like

$$I=\oint_C\left(\frac{1}{\cos(z)-1}\right)dz=7i\int_0^{2\pi}\left(\frac{1}{\cos(7e^{it})-1}\right)e^{it}dt$$

So

$$I=\oint_C\frac{1}{\cos(z)-1}dz=2\pi i\sum_{j=1}^n\operatorname{Res}(f(z),z_j)$$

The integrand has poles of order $1$ whenever the denominator goes to $0$, that is, whenever $\cos(z)=1$ which happens when $z=2\pi n$. How many of these fit within the diameter $14$?

Well, $2\pi\approx6.28<7$ so there are $3$ poles that lie within the region when $n = 0,\pm1$

How do we find the residues? We can write for $z\neq2\pi n$

$$-\frac{1}{1-\cos(z)}=-\sum_{p=0}^\infty\left(\sum_{q=0}^\infty\frac{(-1)^q}{(2q)!}z^{2q}\right)^p$$

In particular, there are never any terms that are of the form $\displaystyle\frac{C_{-1}}{z-z_0}$.

$$\therefore \operatorname{Res}\left(\frac{1}{\cos(z)-1},2\pi n\right)=0$$

And

$$\oint_C\left(\frac{1}{\cos(z)-1)}\right)dz=0$$

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  • $\begingroup$ can you explain this step? 11−cos(z)=−∑p=0∞(∑q=0∞(−1)q(2q)!z2q)p? $\endgroup$ Commented Jun 11, 2018 at 22:44
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    $\begingroup$ Sure, one result is that $\frac{1}{1-x}=\sum_{n=0}^\infty x^n$ given $|x|<1$. In this case, we're using $\cos(x)$ in place of $x$ since $|\cos(x)|<1$ when $x\neq2\pi m$ (and $x\neq\pi m$ which I forgot in the solution) for integer $m$. So $\frac{1}{1-\cos(x)}=\sum_{n=0}^\infty(\cos(x))^n$. Then $\cos(x)$ has its own series representation, $\cos(x)=\sum_{j=0}^\infty\frac{(-1)^jx^{2j}}{(2j)!}$ $\endgroup$
    – mallan
    Commented Jun 11, 2018 at 23:51
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You can use residues to calculate this integral.

The solution would look like this:

  1. Find the when the denominator is equal to $0$. That would lead as to solve the following equation: $$\cos(z) = 1. \tag{1}$$ From the definition of $\cos(z)$ we can easily find out that $z = 2 \pi n, n \in \mathbb{Z}$.
  2. You have to notice that $(\cos(z) - 1)' = 0$ for $z = 2 \pi n$, but $(\cos(z) -1)'' \neq 0$ for $z = 2 \pi n$.
  3. You have to find all the points which are inside your circle $C$.
  4. Know you can calculate the residues using this formula for example: $$Res(f, z) = \frac{1}{(n-1)!} \lim \limits_{z \to z_0} \frac{\mbox{d}^{n-1}}{\mbox{d}z^{n-1}} \big( (z-z_0)^n f(z) \big).$$
  5. Let's make a set of all the $z$ for which the denominator is equal to $0$ and call it $Z$.

  6. Finally you can calculate you integral using this dependence: $$\displaystyle\oint_Cf(z)\mbox{d}z = 2 \pi i \sum \limits_{z \in Z} Res(f, z).$$


In your case $n = 2$, because the second derivative is $\neq 0$ thus we have the second fold of zero.

Done!

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