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Complex Integral

  • Thread starter zenite
  • Start date
  • #1
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1. Integrate z2/(z4-1) counterclockwise around x2 + 16y2=4


2. Cauchy's Integral Forumula


3. Solution
I found the points z=1,-1,i,-i where the function is not defined. Using partial fractions to split them up, and integral them separately.

Only points z=1,-1 lies in the contour, so...
[tex]\oint0.25/(z-1) + 0.25/(z+1) + 1/(z^2+1) dz[/tex]
= 0.25(2Pi I + 2Pi I) + 0 = Pi I

Ans is 0. can anyone find my mistake?



1. Integrate sinh2z/z4 counterclockwise around the unit circle.


2. Cauchy's Integral Forumula


3. Solution

[tex]\oint sinh2z/z^4 = \oint sinh2z/(z-0)^4[/tex]
= 2*PI*i/3! * (sinh2z)'''

Differentiating sinh2z thrice gives 8cosh2z

Hence, integral at z=0 = (8/3)*PI*i

Ans is (8/3)*PI. Again, can anyone spot my mistake.
 

Answers and Replies

  • #2
1,444
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for the first question:

the residue at z=1 is the limit as z goes to 1 of:

[itex](z-1) \left( 0.25 / (z-1) + 0.25 / (z+1) + 1 / (z^2+1) \right)[/itex]
put that in and you get 0.25 + (-0.25) = 0

i imagine this will also happen at the z=-1 pole

then just use Cauchy's residue theorem that

[itex] \int_\gamma f(z) dz = \displaystyle \sum_i Res(f, c_i)[/itex] where [itex]c_i[/itex] are the poles of [itex]f(z)[/itex] and you'll get the whole thing to integrate to 0+0=0
 

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