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

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

My Attempt/Solution:

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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?

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Evaluate by integrating around a suitable closed contour:

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

My Attempt/Solution:

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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 :).

**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 :).

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