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Math Help - Complex line integral

  1. #1
    MHF Contributor arbolis's Avatar
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    Complex line integral

    Calculate I=\oint _{C_R} z^n \log (z)dz for n integer where C_R is the circle with radius R>0 with center in 0 and positively oriented.

    My attempt : (long exercise so I skip a lot of arithmetic) : A parametrization of C_R is Re^{it}.
    z=Re^{it}\Rightarrow dz=iRe^{it}dt.
    \Rightarrow z^n \log (z)=R^ne^{int}[\ln (R)+it] \Rightarrow I=\int _0^{2\pi} i R e^{it}[R^n e^{int}[\ln (R)+it]]dt.
    Skipping steps, I get that I=R^{n+1} i \ln (R) \underbrace{ \int _0^{2\pi} e^{it(n+1)} dt}_{J} - R^{n+1} \underbrace {\int _0^{2\pi} te^{it(n+1)}dt}_{K}.
    Skipping arithmetic once again, I get that J=0 and K=-\frac{2\pi i}{(n+1)}.
    Hence the final result I=\frac{2\pi iR^{n+1}}{(n+1)}.
    I'm more than sure I made many mistakes... can you check my result?
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  2. #2
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    Quote Originally Posted by arbolis View Post
    Calculate I=\oint _{C_R} z^n \log (z)dz for n integer where C_R is the circle with radius R>0 with center in 0 and positively oriented.

    My attempt : (long exercise so I skip a lot of arithmetic) : A parametrization of C_R is Re^{it}.
    z=Re^{it}\Rightarrow dz=iRe^{it}dt.
    \Rightarrow z^n \log (z)=R^ne^{int}[\ln (R)+it] \Rightarrow I=\int _0^{2\pi} i R e^{it}[R^n e^{int}[\ln (R)+it]]dt.
    Skipping steps, I get that I=R^{n+1} i \ln (R) \underbrace{ \int _0^{2\pi} e^{it(n+1)} dt}_{J} - R^{n+1} \underbrace {\int _0^{2\pi} te^{it(n+1)}dt}_{K}.
    Skipping arithmetic once again, I get that J=0 and K=-\frac{2\pi i}{(n+1)}.
    Hence the final result I=\frac{2\pi iR^{n+1}}{(n+1)}.
    I'm more than sure I made many mistakes... can you check my result?
    You should be careful how to parametrize the circle. You see the typical definition for \log z is \log r + i\theta where z = re^{i\theta},r>0,\theta \in (-\pi,\pi]. Thus, parametrize C_R as \gamma: [-\pi,\pi ] \to \mathbb{C} with \gamma(t) = Re^{it}. So the integral will look exactly like before. Just not from 0 to 2\pi but instead -\pi to \pi. However, this is not an issue here because if you replace t by t+\pi and use substitution it does not change the integral.
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  3. #3
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    You should be careful how to parametrize the circle. You see the typical definition for \log z is \log r + i\theta where z = re^{i\theta},r>0,\theta \in (-\pi,\pi]. Thus, parametrize C_R as \gamma: [-\pi,\pi ] \to \mathbb{C} with \gamma(t) = Re^{it}. So the integral will look exactly like before. Just not from 0 to 2\pi but instead -\pi to \pi. However, this is not an issue here because if you replace t by t+\pi and use substitution it does not change the integral.
    Ah I see. You mean I should have took care of the limits of the integral, right?
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    Quote Originally Posted by arbolis View Post
    Ah I see. You mean I should have took care of the limits of the integral, right?
    I am not sure how you defined the complex logarithm. What kind of branch do you use? Most books and most mathematicians define it along the branch (-\infty,0] and so when you compute the logarithm of re^{i\theta} you need to make sure that \theta \in (-\pi,\pi]. But if you defined the branch along the non-negative real axis then then the limits you used are correct.
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    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    I am not sure how you defined the complex logarithm. What kind of branch do you use? Most books and most mathematicians define it along the branch (-\infty,0] and so when you compute the logarithm of re^{i\theta} you need to make sure that \theta \in (-\pi,\pi]. But if you defined the branch along the non-negative real axis then then the limits you used are correct.
    I've just checked my notes, it's defined along the branch (-\infty,0]. Thanks for the clarification.
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