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SkykhanFalcon

  • 2 years ago

\[\int\limits_{C}^{}dz/z\] where c consists of three line segments: From z = 1 to z = 1-i From z = 1-i to z = -1-i From z = -1-i to z = -1

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  1. agentx5
    • 2 years ago
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    At first I though this wasn't going to be so hard, but then I read the sub-instructions >_< @amistre64 , @TuringTest , any insight on this one guys? PS: You did post ALL the contextual information for this one this time right @SkykhanFalcon ? ;-) (/me remembers your last super-challenging differentials question)

  2. agentx5
    • 2 years ago
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    There's no special initial conditions, restrictions, or anything like that right?

  3. SkykhanFalcon
    • 2 years ago
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    :) yeah i remember, for now there is not :)

  4. amistre64
    • 2 years ago
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    this appears to be the same question that was posted earlier

  5. SkykhanFalcon
    • 2 years ago
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    @experimentX dude i could not do that :/

  6. SkykhanFalcon
    • 2 years ago
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    and today is the last day :/

  7. experimentX
    • 2 years ago
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    which one??

  8. SkykhanFalcon
    • 2 years ago
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    they are all connected

  9. experimentX
    • 2 years ago
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    i think i already did the first one.

  10. SkykhanFalcon
    • 2 years ago
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    you found sqrt2-i(pi)/4

  11. SkykhanFalcon
    • 2 years ago
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    i am not sure it is true or not

  12. experimentX
    • 2 years ago
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    what's the answer supposed to be??

  13. SkykhanFalcon
    • 2 years ago
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    final answer should be -(pi)*i

  14. experimentX
    • 2 years ago
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    for question no 1??

  15. SkykhanFalcon
    • 2 years ago
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    no they are connected each other book says only an answer and it is -(pi)*i

  16. experimentX
    • 2 years ago
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    O ... let's see for 2

  17. experimentX
    • 2 years ago
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    |dw:1343076384065:dw|

  18. experimentX
    • 2 years ago
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    |dw:1343076480604:dw|

  19. experimentX
    • 2 years ago
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    |dw:1343076552517:dw|

  20. experimentX
    • 2 years ago
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    |dw:1343076614227:dw|

  21. experimentX
    • 2 years ago
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    |dw:1343076688932:dw| sorry man ... i couldn't get any better

  22. SkykhanFalcon
    • 2 years ago
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    :) okey bro thanks again

  23. SkykhanFalcon
    • 2 years ago
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    @mukushla @richyw @eliassaab any idea ?

  24. eliassaab
    • 2 years ago
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    Have you had the residue theorem?

  25. SkykhanFalcon
    • 2 years ago
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    yes i guess

  26. SkykhanFalcon
    • 2 years ago
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    but i did not learn well :/

  27. eliassaab
    • 2 years ago
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    If you know the Residue Theorem you can cut your work from computing three line integrals to one.

  28. eliassaab
    • 2 years ago
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    Your line integral on the first line is z = 1 +i y dz = i dy \[ \int_0^{-1} \frac{i}{1+i y} \, dy=\frac{1}{4} (\ln(4)-i \pi ) \]

  29. eliassaab
    • 2 years ago
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    Mimic my post above for 2 remaining lines.

  30. SkykhanFalcon
    • 2 years ago
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    hmm then will i continue with 1/4(ln4.... or i always use the i/1+iy

  31. SkykhanFalcon
    • 2 years ago
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    @eliassaab ?

  32. eliassaab
    • 2 years ago
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    for the second integral z= x -I dz = dx \[\int_1^{-1} \frac{1}{x-i} \, dx=-\frac{1}{2} (i \pi ) \]

  33. SkykhanFalcon
    • 2 years ago
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    how can i understand the integral part for ex. why the second one become 1/x-i

  34. eliassaab
    • 2 years ago
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    How much is y on the second line? -1 How much is x on the seconfd line? x from 1 to -1 z= x -i

  35. eliassaab
    • 2 years ago
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    On the third line z= -1 + i y dz= i dy \[ \int_{-1}^0 \frac{i}{-1+i y} \, dy \]

  36. SkykhanFalcon
    • 2 years ago
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    ok i understand i guess thank u :)

  37. eliassaab
    • 2 years ago
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    \[ \int_{-1}^0 \frac{i}{-1+i y} \, dy=-\frac{1}{4} i (\pi -2 i \log (2)) \]

  38. eliassaab
    • 2 years ago
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    log=ln

  39. SkykhanFalcon
    • 2 years ago
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    thank u so much :)

  40. eliassaab
    • 2 years ago
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    yw

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