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Math Help - Contour Integration

  1. #1
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    Contour Integration

    f(z) = \frac{z+2}{2 } evaulate \int c f(z) dz when C is the semicircle z=2{e}^{ i\theta } , 0 \leqslant \theta  \leqslant  \pi

    I am not getting what my lecturer gets, ill show you what ive done.
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  2. #2
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    Quote Originally Posted by adam_leeds View Post
    f(z) = \frac{z+2}{2 } evaulate \int c f(z) dz when C is the semicircle z=2{e}^{ i\theta } , 0 \leqslant \theta  \leqslant  \pi

    I am not getting what my lecturer gets, ill show you what ive done.
    My Attempt

    dz = 2i{e}^{ i\theta } d\theta

    and then putting this in and integrating it betwen pi and 0, is this right?

    If so what are other people getting for their final answer?
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  3. #3
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    I get -4. What do you get?
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  4. #4
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    Quote Originally Posted by Ackbeet View Post
    I get -4. What do you get?
    Not the same

    \int \frac{{2e}^{ i\theta } + 2}{2 } 2i{e}^{ i\theta } d\theta = \frac{4i{e}^{ 2i\theta } + 4i{e}^{ i\theta } }{2 } d\theta = \int 2i{e}^{ 2i\theta } +2i{e}^{ i\theta }d\theta

    Is that right?
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  5. #5
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    =2i\int_{0}^{\pi}(e^{2i\theta}+e^{i\theta})\,d \theta.

    That is what I get, which agrees with your work so far. What next?
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  6. #6
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    = {e}^{ 2i\theta } + {2e}^{ 2i\theta } between \pi and  0 so {e}^{ 2i\pi } + {2e}^{ 2i\pi} - 3
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  7. #7
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    Your antiderivative is incorrect:

    2i\int_{0}^{\pi}(e^{2i\theta}+e^{i\theta})\, d\theta=2i\left(\frac{e^{2i\theta}}{2i}+\frac{e^{i  \theta}}{i}\right)\Bigg|_{0}^{\pi}.
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  8. #8
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    Quote Originally Posted by Ackbeet View Post
    Your antiderivative is incorrect:

    2i\int_{0}^{\pi}(e^{2i\theta}+e^{i\theta})\, d\theta=2i\left(\frac{e^{2i\theta}}{2i}+\frac{e^{i  \theta}}{i}\right)\Bigg|_{0}^{\pi}.
    How does that make it -4 putting the limits in?
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  9. #9
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    Well, what do you get?
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  10. #10
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    Quote Originally Posted by Ackbeet View Post
    Well, what do you get?
    {e}^{2i\pi  } + 2{e}^{i\pi  } -3
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  11. #11
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    Quote Originally Posted by adam_leeds View Post
    {e}^{2i\pi  } + 2{e}^{i\pi  } -3
    =1-2-3\dots
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  12. #12
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    Quote Originally Posted by Ackbeet View Post
    =1-2-3\dots
    Thanks but how?
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  13. #13
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    Well,

    e^{i\pi}+1=0, one of the most beautiful equations in all of mathematics, and also

    e^{2i\pi}=1.

    Both are points on the unit circle in the complex plane. Another way of thinking about this is that

    e^{i\theta}=\cos(\theta)+i\sin(\theta). Hence,

    e^{i\pi}=\cos(\pi)+i\sin(\pi)=-1+i0=-1, and similarly for e^{2i\pi}.

    Make sense?
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  14. #14
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    Quote Originally Posted by Ackbeet View Post
    Well,

    e^{i\pi}+1=0, one of the most beautiful equations in all of mathematics, and also

    e^{2i\pi}=1.

    Both are points on the unit circle in the complex plane. Another way of thinking about this is that

    e^{i\theta}=\cos(\theta)+i\sin(\theta). Hence,

    e^{i\pi}=\cos(\pi)+i\sin(\pi)=-1+i0=-1, and similarly for e^{2i\pi}.

    Make sense?
    Yep, thanks.
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  15. #15
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    You're welcome!
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