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3. Evaluate each of the following contour integrals with the Residue Theorem. (a) \( \int_{|z|=4} \frac{z+1}{z(z-2)} dz \) (b) \( \int_{|z|=2} \frac{dz}{z^3(z+4)} \) (c) \( \int_{|z|=4} \frac{dz}{z \sin z} \) (d) \( \int_{|z|=3} (z-1)e^{3(z-1)^{-2}} dz \)

          3. Evaluate each of the following contour integrals with the Residue Theorem.
(a) \( \int_{|z|=4} \frac{z+1}{z(z-2)} dz \)
(b) \( \int_{|z|=2} \frac{dz}{z^3(z+4)} \)
(c) \( \int_{|z|=4} \frac{dz}{z \sin z} \)
(d) \( \int_{|z|=3} (z-1)e^{3(z-1)^{-2}} dz \)
        
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3. Evaluate each of the following contour integrals with the Residue Theorem.
(a) ∫|z|=4(z+1)/(z(z-2)) dz
(b) ∫|z|=2(dz)/(z^3(z+4))
(c) ∫|z|=4(dz)/(z sin z)
(d) ∫|z|=3 (z-1)e^3(z-1)^-2 dz

Added by Mark M.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Evaluate each of the following contour integrals with the Residue Theorem: (a) ∮(z+1)/(z^2+4z-2) dz (b) ∮(1)/(23z+4) dz (c) ∮(4zsin^2(z-1))/(e^3-1-d) dz
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Transcript

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00:01 Hello students, we need to evaluate the following using residue theorem.
00:05 Given integral over the boundary c 3z plus 2 whole square divided by z into z minus 1 into 2z plus 5 dz.
00:19 Given the condition mod z, that is boundary c, is equal to 3.
00:25 Now here we can observe that we have alpha is equal to 0, 1 and minus 5 by 0.
00:32 Simple poles of f of z is given by that is f of z we have the value 3 z plus two whole square divided by z into z minus 1 into 2 z plus 5 now residue of z comma z is equal to 0 we have limit z 10 into 0 into z minus 0 into f of z this implies we have limit z ending to 0 3 z plus 2 whole square divided by z into z minus 1 into 2 z plus 5.
01:16 This implies we have 4 by minus 1 into 5 which gives the value minus 4 by 5.
01:23 Now residue of f of z that is z is equal to 1 we have 3 into 1 plus 2 whole square divided by 1 1 1 1 2 into 1 plus 5 which gives the value 25 by 7 and residue of f of z when z is equal to minus 5 by 2 we have minus 15 by 2 plus 2 whole square divided by minus 5 by 2 into minus 5 by 2 minus 5 by 2 minus 1 which gives the value 12 by 21 by 35 therefore integral over the boundary we see f z into d z is equal to 2 pi i into summation of residue of f of z into where z belongs to alpha.
02:18 This implies we have 2 pi i into minus 4 by 5 plus 25 by 7 plus 121 by 35.
02:29 On calculating and simplifying, therefore we get the value 436 pi i divided by 35...
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