complex analysis4

**shaheen7** · May 4th 2011, 12:40 PM

a. Evaluate the integral
\int e^z/((z^2+pi)^2) where the integral is the circle of radius 4 centered at the origin.

b. Evaluate the integral
\int 1/((z^2+1)(z^2-1))
where the integral is the circle of radius 2 centered at the origin.

**TheEmptySet** · May 4th 2011, 01:32 PM

Originally Posted by shaheen7

a. Evaluate the integral
\int e^z/((z^2+pi)^2) where the integral is the circle of radius 4 centered at the origin.

b. Evaluate the integral
\int 1/((z^2+1)(z^2-1))
where the integral is the circle of radius 2 centered at the origin.

Use the residue theorem

$\oint \frac{e^z}{(z^2+\pi^2)^2}dz=\oint \frac{e^z}{(z+i\pi)^2(z-i\pi)^2}dz$

So the function has two poles of order 2.

So we need to calculate the residues at
$\pm i\pi$

So to calculate at

$i\pi$

we have

$\text{Res}(f,i\pi) = \lim_{z \to i\pi }\frac{d}{dz}(z-i\pi)^2\left(\frac{e^z}{(z+i\pi)^2(z-i\pi)^2} \right)$

$\lim_{z \to i\pi} \frac{d}{dz}\frac{e^{z}}{(z+i\pi)^2}= \lim_{z \to i\pi} \frac{e^z(z+i\pi)^2-2e^{z}(z + i \pi)}{(z+ i\pi )^4}$

$\frac{e^{i\pi}(2i\pi)(2i\pi-2)}{(2i\pi)^4}=\frac{-1(2 i \pi -2)}{(2i \pi)^3 }=\frac{2-2\pi i}{-8 i\pi^3}=\frac{1}{4}\left(\frac{\pi +i}{\pi^3} \right)$

Now you compute the other one and remember that the answer is

$2 \pi i \sum \text{Res}(f)$

at all of the poles in the contour.

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