hi all, my first post; had a minor headache with this problem lol.(adsbygoogle = window.adsbygoogle || []).push({});

PROBLEM 1:

Finding Residue:

-----------------

find Res(g,0) for [itex] g(z) = z^{-2}coshz[/itex]

My Attempt/Solution:

-----------------

I know [itex] coshz = 1 + \frac{x^2}{2!} + \frac{z^4}{4!} .... [/itex]

so now [itex] 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!}[/itex]

we know the residue is the coefficient of the -1th term (or the coefficient of [itex]z^{-1}[/itex]) 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:

----------------------------

Evaluate by integrating around a suitable closed contour:

[itex]\int_{-\infty}^{\infty} \frac{cos3x}{x^2 + 4}[/itex]

My Attempt/Solution:

-----------------

First consider the intergral,

[itex] I_R = \int_{\gamma^R} \frac{e^{3iz}}{(z+2i)(z-2i)}dz[/itex] where [itex] \gamma^R = \gamma^R_1 + \gamma^R_2 [/itex] and [itex]\gamma^R_1 = \{|z| = R , Im(z) > 0\} [/itex] and [itex]\gamma^R_2 = \{z = Re(z) , -R\leq z \leq R\}[/itex]

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

[itex]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}[/itex]

By Cauchy's Residue Theorem, we have:

[itex] I_R = (2\pi i)\frac{e^{-6}}{4i} = \frac{\pi e^{-6}}{2}[/itex]

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

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

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

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Residue Problem.

**Physics Forums | Science Articles, Homework Help, Discussion**