Residue at poles of complex function

In summary, the conversation is about finding the correct factor to use in the residue formula in order to solve a problem involving poles and the residue theorem. The correct factor is not ##(3z+2)^3## as originally thought, but rather ##(z-a)^3## where ##a=(-2/3)##. The conversation also touches on a solved example where the factor was incorrectly broken down.
  • #1
jaus tail
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Homework Statement


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Homework Equations


First find poles and then use residue theorem.

The Attempt at a Solution


upload_2018-2-2_11-33-42.png

Book answer is A. But there's no way I'm getting A. The 81 in numerator doesn't cancel off.
 

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  • #2
jaus tail said:

Homework Statement


View attachment 219529

Homework Equations


First find poles and then use residue theorem.

The Attempt at a Solution


View attachment 219528
Book answer is A. But there's no way I'm getting A. The 81 in numerator doesn't cancel off.

Be careful. You don't multiply the function by ##(3z+2)^3## before you differentiate it. What's the correct factor? You won't get the books answer A in any case, but that's the books problem.
 
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  • #3
I don't get your answer or any of the answers above. It's hard for me to read your writing.
 
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  • #4
Dick said:
Be careful. You don't multiply the function by ##(3z+2)^3## before you differentiate it. What's the correct factor? You won't get the books answer A in any case, but that's the books problem.
The (3z+2)3 will get canceled when I find residue at z = -2/3, just like the (3z-2)3 part will get canceled when I find residue at z = 2/3.
 
  • #5
jaus tail said:
The (3z+2)3 will get canceled when I find residue at z = -2/3, just like the (3z-2)3 part will get canceled when I find residue at z = 2/3.

The residue formula has ##(z-a)^3## where ##a=(-2/3)##. That's not ##(3z+2)^3##.
 
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  • #6
upload_2018-2-2_20-44-35.png

Yeah you're right. In solved example they've broken 0.5z - 1.5i to z - 3j and multiplied 2 into numerator. Thanks.
 

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1. What is a residue at a pole of a complex function?

A residue at a pole of a complex function is a complex number that represents the value of the function at that particular pole. It is calculated using the Cauchy residue theorem and is important in evaluating integrals of complex functions.

2. How do you calculate the residue at a simple pole?

To calculate the residue at a simple pole, you can use the formula Res(f,c) = lim(z→c) [ (z-c) * f(z) ], where c is the location of the pole and f(z) is the complex function. You can also use the Laurent series expansion to find the coefficient of the term with (z-c)^-1.

3. Can a complex function have more than one pole?

Yes, a complex function can have multiple poles. The number of poles a function has is determined by the degree of the denominator of the function's rational expression. For example, a function with a denominator of (z-1)^3 will have three poles at z=1.

4. What happens to the residue at a pole if the function has a removable singularity?

If a function has a removable singularity at a particular pole, then the residue at that pole is equal to 0. This is because the function is continuous at that point and can be extended to that point without any singularity.

5. Why are residues important in complex analysis?

Residues play a crucial role in evaluating contour integrals of complex functions. They allow us to calculate integrals that would be difficult or impossible to evaluate using traditional methods. They are also used in solving differential equations and in determining the behavior of functions near their poles.

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