Residue at poles of complex function

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The discussion centers around finding residues at the poles of a complex function using the residue theorem. Participants express confusion over the correct application of the theorem, particularly regarding the cancellation of factors in the numerator. There is a specific mention of the factor (3z+2)^3 and its role in calculating residues at z = -2/3 and z = 2/3. The conversation highlights discrepancies between personal calculations and the book's answer, emphasizing the need for careful differentiation and factor handling. Overall, the thread illustrates the complexities involved in applying the residue theorem correctly.
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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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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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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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.
 
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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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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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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