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I have to find the value of:

$\int_{|z-2|=\frac{3}{2}}\frac{\cos{z}}{z^2(z^2-\pi^2)}$

If there is no singularity in the closed disk $D=|z-2|\leq\frac{3}{2}$ then the integral should be 0. For the points where there is a singularity in $D$ I think using residues would work, but considering where I got this from it should be possible to do it without, probably using cauchy's integral formula, but I can't see how.

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  • $\begingroup$ what if $z=\pi$? because $\pi-2$ falls within 1.5 $\endgroup$
    – imranfat
    Apr 9, 2015 at 18:59

2 Answers 2

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Put $g(z)=cos(z)/[z^2(z^2-\pi^2)]$, note that $$ g(z)= \frac{cos(z)}{z^2(z-\pi)(z+\pi)} = \frac{f(z)}{z-\pi} $$ were $f(z)=cos(z)/[z^2(z+\pi)]$. Since $g$ has a singularity in $z_0=\pi$, and clearly $\pi \in D^{o}$, cause $|\pi -2|< 1.5$. So $g$ is analytic on $D \backslash \left\{ \pi \right\}$, then by Cauchy I. F. we have $$ \int_{\partial D} g(z) dz = \int_{\partial D}\frac{f(z)}{z-\pi} dz = 2\pi i f(\pi) = 2\pi i \left(\frac{cos(\pi)}{\pi^2(\pi+\pi)}\right)=-\frac{i}{\pi^2} $$

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The function $f(z) = \cos(z)/[z^2(z + \pi)]$ is analytic in $D$ and $\pi$ lies in the interior of $D$. Hence, by the Cauchy integral formula, the value of your integral is

$$2\pi i f(\pi) = \frac{-2\pi i}{\pi^2(2\pi)} = -\frac{i}{\pi^2}.$$

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  • $\begingroup$ Edited the question for clarity, are you sure $f$ is analytic on every $D_z$? For example when $\pi \in D$ I don't think it is. $\endgroup$
    – mikos
    Apr 9, 2015 at 19:13
  • $\begingroup$ What do you mean "on every D_z"? The disc is centered at $2$ and has radius $3/2$. Certainly, there's only one such disc, and you're integrating over the boundary of that disc. $\endgroup$
    – kobe
    Apr 9, 2015 at 19:14
  • $\begingroup$ my bad I got mixed up $\endgroup$
    – mikos
    Apr 9, 2015 at 19:21
  • $\begingroup$ @mikos ok, so do you follow now? $\endgroup$
    – kobe
    Apr 9, 2015 at 19:33
  • $\begingroup$ Yes, I got it now. Thanks. $\endgroup$
    – mikos
    Apr 9, 2015 at 21:54

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