5.38 Evaluate (e^z)/((z^2 + π^2)^2) dz where C is the circle...

5.38 Evaluate \( \oint_C \frac{e^z}{(z^2 + \pi^2)^2} dz \) where C is the circle \(|z| = 4\). Solution The poles of \( \frac{e^z}{(z^2 + \pi^2)^2} = \frac{e^z}{(z - \pi i)^2 (z + \pi i)^2} \) are at \( z = \pm \pi i \) inside C and are both of order two. Residue at \( z = \pi i \) is \( \lim_{z \to \pi i} \frac{1}{1!} \frac{d}{dz} \left\{ (z - \pi i)^2 \frac{e^z}{(z - \pi i)^2 (z + \pi i)^2} \right\} = \frac{\pi + i}{4\pi^3} \). Residue at \( z = -\pi i \) is \( \lim_{z \to -\pi i} \frac{1}{1!} \frac{d}{dz} \left\{ (z + \pi i)^2 \frac{e^z}{(z - \pi i)^2 (z + \pi i)^2} \right\} = \frac{\pi - i}{4\pi^3} \). Then \( \oint_C \frac{e^z}{(z^2 + \pi^2)^2} dz = 2\pi i \) (sum of residues) \( = 2\pi i \left( \frac{\pi + i}{4\pi^3} + \frac{\pi - i}{4\pi^3} \right) = \frac{i}{\pi} \)

Transcript

00:01 Hello everyone this question we have to find the residue at z equal to 0.

00:04 So the first part can be written as 1 by z into 1 plus z which will be nothing but 1 by z into 1 plus z.

00:11 So this implies we can see simple pole at z equal to 0.

00:18 Therefore the residue is the coefficient of 1 by z term in the lorentz series.

00:34 So which is 1.

00:37 Therefore a residue at z equal to 0 is 1.

00:45 Then the second part is z cos 1 by z.

00:49 So z cos by 1 is z by using lorentz expansion we can write 1 minus 1 by z the whole square by 2 factorial plus 1 by z the whole power 4 by 4 factorial minus 1 by z to the power 6 by 6 factorial.

01:09 So this can be written as z minus z by 2 factorial z by 4 factorial minus z by 6 factorial.

01:18 So the residue is the coefficient of 1 by z in the lorentz series.

01:26 So which is the residue.

01:29 So here it is minus 1 by 2 factorial plus 1 by 4 factorial minus 1 by 6 factorial which will give you minus 1 by 2 plus 1 by 24 minus 1 by 720 goes on it goes on.

01:43 So if you simplify we will get residue as minus 1 by 2 or minus 0 .5.

01:52 So the third part is z minus sine z by z.

01:54 So if you see this f of z as a removable singularity at z equal to 0 and sine z by 2 as element 1.

02:02 So we can extend this f of s as continuous function at z equal to 0 by defining f of 0 equal to 1.

02:09 So sine z expansion is this.

02:10 So z minus z by z if you see this we will be having the by simplifying we will be having the coefficient of z minus 0 to the power minus 1 is 0.

02:23 So the residue is 0.

02:25 Then the d part is cot z by z to the power 4.

02:30 So your cot z can be written as sine z divided by cos z.

02:34 So which is nothing but 1 minus z square by 2 factorial z to the power by 4 factorial plus order of z to the power 6.

02:42 Here it will become z minus z square by 2 factorial z cube by 5 factorial.

02:50 So it is z cube by 3 factorial z to the power 5 factorial plus o of z to the power 7.

02:57 So further simplifying so if you see here i am simplifying like this i am getting the coefficient of laurent series at z equal to 0 will be minus 145.

03:08 So the residue is minus 145.

03:12 So similarly the last part is sine hz divided by z for 1 minus z square.

03:17 I factorize the denominator like this from this we can get simple pole at z equal to 0.

03:22 So i use laurent series expansion of sine hz...

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