Find the Residue for sin(z)/z^4

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In summary, the conversation discusses the calculation of the residue of sin(z)/z^4 using the formula for a pole of order m at 0. However, there is a mistake in taking the second derivative of sin(z)/z, which leads to incorrect results. It is suggested to use l'Hospital's Rule to find the correct value. Another simpler method is to use the Laurent series for sin(z) and determine the coefficient of z^-1, which yields the correct residue of -1/6.
  • #1
zezima1
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Homework Statement


Calculate the residue of sinz/z^4

Homework Equations


The residue for a pole of order m at 0 is given my:
lim z->0 [1/(m-1)!dm-1/dzm-1[(zmf(z)]]


The Attempt at a Solution


Clearly the pole has order 3:
So we get:
Re(0) = limz->0[1/2d2/dz2[sinz/z]]
I get that the only term in the limit that doesn't go to zero is:
Re(0) = limz->0[-1/2sinz/z] = -1/2
But I should get -1/3!
Apparantly something goes wrong somewhere as I should get a tree from differentiation. Can anyone see where that is? :S
 
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  • #2
You didn't take the second derivative of sin(z)/z correctly
 
  • #3
hmm I thought I might have done something wrong there. Let's go over it:

(sinz/z)'' = (cosz/z-sinz/z2)' = -sinz/z - 2cosz/z2+3sinz/z3

Now as we let z->0 the only term that survives is sinz/z which ->-1. Where is my mistake?
 
  • #4
zezima1 said:
hmm I thought I might have done something wrong there. Let's go over it:

(sinz/z)'' = (cosz/z-sinz/z2)' = -sinz/z - 2cosz/z2+3sinz/z3

Now as we let z->0 the only term that survives is sinz/z which ->-1. Where is my mistake?

Your last term should be ##\frac{2 \sin{(z)}}{z^3}##. Go through that derivative again to find your mistake.

Also, the other terms do approach values, they don't "disappear." Use l'Hospital's Rule.
 
  • #5
okay yes I see what I did wrong now. I accidently took the terms involving z of a higher order than one to go to zero when they really diverge.
 
  • #6
aaaa202 said:
okay yes I see what I did wrong now. I accidently took the terms involving z of a higher order than one to go to zero when they really diverge.

Well, no, they don't diverge. They converge to specific points (like I said, use l'Hospital's Rule). Also, you didn't take the second derivative properly which would yield bad results as well.
 
  • #7
Are you required to use that formula? Often the simplest way to find the residue of a function, at a given value of z, is to use the fact that the residue is the coefficient of [itex]z^{-1}[/itex] in the Laurent series for f(z) about the given value.

Here, we can start with the Taylor's series for sin(z): [itex]z- (1/6)z^3+ \cdot\cdot\cdot[/itex] so that
[tex]\frac{sin(z)}{z^4}= \frac{1}{z^3}- \frac{1}{6z}+ \cdot\cdot\cdot[/tex]
so that the residue is, as you say, -1/6.
 

FAQ: Find the Residue for sin(z)/z^4

What is the meaning of "Find residue of sinz/z^4"?

The "residue" in this context refers to the coefficient of the term with the highest negative power in the Laurent series expansion of the function sin(z)/z^4. In simpler terms, it is the value of the term with the smallest exponent when the function is written in a series form.

Why is it important to find the residue of a function?

The residue of a function is important in complex analysis because it helps in calculating integrals and evaluating the behavior of a function at singular points. It also has applications in physics, engineering, and other fields.

How do you find the residue of sinz/z^4?

To find the residue of sinz/z^4, you can use the formula Res(f, z0) = lim(z->z0) [(z-z0)^n * f(z)], where n is the order of the pole at z0. In this case, since z^4 is a fourth-order pole, n=3. So, the residue can be calculated as Res(sin(z)/z^4, 0) = lim(z->0) [(z-0)^3 * sin(z)/z^4] = lim(z->0) z^2 * sin(z)/z^4 = lim(z->0) sin(z)/z^2 = 1.

Is it possible to find the residue at points other than the origin?

Yes, it is possible to find the residue of a function at any point where it has a singularity, such as a pole or a branch point. However, the formula for calculating the residue may vary depending on the order and type of singularity at that point.

Can the residue of a function be negative?

Yes, the residue of a function can be negative. The sign of the residue depends on the direction of the contour used for integration. If the contour is traversed in a clockwise direction, the residue will be negative, and if it is traversed in a counterclockwise direction, the residue will be positive.

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