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Integration using residue theorem

  1. Jul 23, 2014 #1
    Hi. I have to use the residue theorem to integrate f(z).
    Can someone help me out? I am stuck on the factorization part.

    Find the integral
    $$\int_{0}^{2\pi} \,\frac{d\theta}{25-24\cos\left({\theta}\right)}$$

    My answer:
    $$\int_{0}^{2\pi} \,\frac{d\theta}{25-24\cos\left({\theta}\right)}=\oint_{c}^{} \,\frac{dz/iz}{25-24(\frac{1}{2}(z+\frac{1}{z}))}$$

    $$=\frac{1}{i}\oint_{c}^{} \,\frac{dz}{-12z^2+25z-12}$$
     
  2. jcsd
  3. Jul 23, 2014 #2

    vela

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    Try something like ##-12z^2 + 25z -12 = -(12z^2-25z+12) = -(3z+\cdots)(4z+\cdots)##.
     
  4. Jul 23, 2014 #3
    Got it.

    Here's my continuation:

    $$\frac{1}{i}\oint_{c}^{}\frac{dz}{-(4z-3)(3z-4)}=-\frac{1}{i}\oint_{c}^{}\frac{dz}{(4z-3)(3z-4)}$$

    It has a singularity at z=3/4

    $$Resf(z)_{|z=\frac{3}{4}|}=\lim_{{z}\to{\frac{3}{4}}}(z-\frac{3}{4})(\frac{1}{(4z-3)(3z-4)})=\frac{-7}{5}$$

    By residue theorem, the integral becomes
    $$2\pi{i}\frac{-1}{i}\frac{-7}{5}=\frac{14\pi}{5} $$

    Is this correct? Check please.
     
  5. Jul 23, 2014 #4
    I had a miscalculation. The final answer should be ##2\pi/7##.
     
  6. Jul 24, 2014 #5

    vela

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    Looks good.
     
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