# Residue Problem.

1. Jul 7, 2008

### mathfied

hi all, my first post; had a minor headache with this problem lol.

PROBLEM 1:
Finding Residue:
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find Res(g,0) for $g(z) = z^{-2}coshz$

My Attempt/Solution:
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I know $coshz = 1 + \frac{x^2}{2!} + \frac{z^4}{4!} ....$

so now $z^{-2}coshz = z^{-2} (1 + \frac{z^2}{2!} + \frac{z^4}{4!} ... ) = \frac{1}{z^2} + \frac{z^2}{2!z^{2}} + \frac{z^4}{4!z^{2}} ... = \frac{1}{z^2} + \frac{1}{2!} + \frac{z^2}{4!}$

we know the residue is the coefficient of the -1th term (or the coefficient of $z^{-1}$) but there is no -1th term as you can see. So does that mean the residue is 0, or am I missing something?

PROBLEM 2:
Finding Integral:
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Evaluate by integrating around a suitable closed contour:
$\int_{-\infty}^{\infty} \frac{cos3x}{x^2 + 4}$

My Attempt/Solution:
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First consider the intergral,
$I_R = \int_{\gamma^R} \frac{e^{3iz}}{(z+2i)(z-2i)}dz$ where $\gamma^R = \gamma^R_1 + \gamma^R_2$ and $\gamma^R_1 = \{|z| = R , Im(z) > 0\}$ and $\gamma^R_2 = \{z = Re(z) , -R\leq z \leq R\}$

Now, let $g(z) = \frac{e^{3iz}}{(z+2i)(z-2i)}$. The integrand has 2 simple poles, one at 2i and one at -2i. Only 2i is inside the contour so:

$Res(g,2i) = \lim_{z \to 2i}\frac{(z-2i)e^{3iz}}{(z+2i)(z-2i)} = \lim_{z \to 2i}\frac{e^{3iz}}{(z+2i)} = \frac{e^{-6}}{4i}$

By Cauchy's Residue Theorem, we have:
$I_R = (2\pi i)\frac{e^{-6}}{4i} = \frac{\pi e^{-6}}{2}$

To show that $\int_{\gamma^R_1}g(z)dz \to 0, R \to \infty$, we apply Jordan's Lemma : $M(R) \leq \frac{1}{R^2-4}$.

So now this means, $\int_{-\infty}^{\infty} \frac{cos3x}{x^2 + 4} = \frac{\pi e^{-6}}{2}$

Please could you verify whether I have worked these out correctly. Thanks a great deal, highly appreciated :).

Last edited: Jul 7, 2008
2. Jul 7, 2008

### Pacopag

Both look good to me.

3. Jul 8, 2008

### HallsofIvy

Staff Emeritus
In problem 1, the fact that " no -1th term" because its coefficient is 0.