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Math Help - Complex Integration

  1. #1
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    Complex Integration

    How would I compute:

    I=\int_{2\pi }^{0}\frac{4d\theta }{2 cos\theta + 3 sin\theta + 2i}

    I have tried finding the poles by changing the values into complex values

    where z=e^{i\theta }

    2cos\theta =z+\frac{1}{z}

    2sin\theta =-i({z-\frac{1}{z}})

    I have concluded

    I=\oint \frac{\frac{4dz}{iz}}{(z+\frac{1}{z})-\frac{3i}{2}(z-\frac{1}{z})+2i}

    Could someone tell me if this correct and I cant find the poles cause I cant find anyway of factorising! Thanks in advance!
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  2. #2
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    Quote Originally Posted by zizou1089 View Post
    How would I compute:

    I=\int_{2\pi }^{0}\frac{4d\theta }{2 cos\theta + 3 sin\theta + 2i}

    => I= - \int_{0}^{2\pi}\frac{4d\theta }{2 cos\theta + 3 sin\theta + 2i}
    I have tried finding the poles by changing the values into complex values

    where z=e^{i\theta }

    2cos\theta =z+\frac{1}{z}

    2sin\theta =-i({z-\frac{1}{z}})

    I have concluded

    I=-\oint \frac{\frac{4dz}{iz}}{(z+\frac{1}{z})-\frac{3i}{2}(z-\frac{1}{z})+2i}

    Could someone tell me if this correct and I cant find the poles cause I cant find anyway of factorising! Thanks in advance!
    Multiply iz with the denominator.

    Simplifying will give us the denominator's expression as,

    ( 3 + 2i ) z^2 - 4 z + ( 2i - 3 )

    Now you should be able to figure out the compex roots quite easily.
    Last edited by ques; May 5th 2010 at 03:22 PM.
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  3. #3
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    Quote Originally Posted by ques View Post
    Multiply iz with the denominator.

    Simplifying will give us the denominator's expression as,

    ( 3 + 2i ) z^2 - 4 z + ( 2i - 3 ) = 0

    Now you should be able to figure out the compex roots quite easily.
    Thats the bit Im stuck on ..
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by zizou1089 View Post
    Thats the bit Im stuck on ..
    Do you know the quadratic formula?
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  5. #5
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    Consider ( 3 + 2i ) z^2 - 4 z + ( 2i - 3 ) = 0

    For a(z^2) + bz + c = 0 the roots are

    z = {-b +- (b^2 - 4ac)^(1/2) }/2a

    Here z = [ 4 +- {16 - ( 2i - 3 ) (2i + 3) }]/ {2(2i + 3)}

    You should be able to take it on from here.
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  6. #6
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    Quote Originally Posted by Bruno J. View Post
    Do you know the quadratic formula?
    I am new here and I am sorry that I gave out way too much!
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  7. #7
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    Yep got it with the quadratic formula, problem is sometimes using the formula i get a complex number within the square root, and the function cannot be factorised either, so is there a more consistent way of getting the roots??
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  8. #8
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    Quote Originally Posted by zizou1089 View Post
    Yep got it with the quadratic formula, problem is sometimes using the formula i get a complex number within the square root, and the function cannot be factorised either, so is there a more consistent way of getting the roots??

    Why do you think that i^(1/5) or that ( i + 1)^(1/35) does not exist ?
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  9. #9
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by ques View Post
    I am new here and I am sorry that I gave out way too much!
    No worries mate, welcome aboard!
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