Residue at poles of complex function

jaus tail
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Homework Statement


upload_2018-2-2_11-34-2.png


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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Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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