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Thread: Complex line integral

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

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

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

    My attempt : (long exercise so I skip a lot of arithmetic) : A parametrization of $\displaystyle C_R$ is $\displaystyle Re^{it}$.
    $\displaystyle z=Re^{it}\Rightarrow dz=iRe^{it}dt$.
    $\displaystyle \Rightarrow z^n \log (z)=R^ne^{int}[\ln (R)+it] \Rightarrow I=\int _0^{2\pi} i R e^{it}[R^n e^{int}[\ln (R)+it]]dt$.
    Skipping steps, I get that $\displaystyle I=R^{n+1} i \ln (R) \underbrace{ \int _0^{2\pi} e^{it(n+1)} dt}_{J} - R^{n+1} \underbrace {\int _0^{2\pi} te^{it(n+1)}dt}_{K}$.
    Skipping arithmetic once again, I get that $\displaystyle J=0$ and $\displaystyle K=-\frac{2\pi i}{(n+1)}$.
    Hence the final result $\displaystyle I=\frac{2\pi iR^{n+1}}{(n+1)}$.
    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 $\displaystyle \log z$ is $\displaystyle \log r + i\theta$ where $\displaystyle z = re^{i\theta},r>0,\theta \in (-\pi,\pi]$. Thus, parametrize $\displaystyle C_R$ as $\displaystyle \gamma: [-\pi,\pi ] \to \mathbb{C}$ with $\displaystyle \gamma(t) = Re^{it}$. So the integral will look exactly like before. Just not from $\displaystyle 0$ to $\displaystyle 2\pi$ but instead $\displaystyle -\pi$ to $\displaystyle \pi$. However, this is not an issue here because if you replace $\displaystyle t$ by $\displaystyle t+\pi$ and use substitution it does not change the integral.
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    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 $\displaystyle \log z$ is $\displaystyle \log r + i\theta$ where $\displaystyle z = re^{i\theta},r>0,\theta \in (-\pi,\pi]$. Thus, parametrize $\displaystyle C_R$ as $\displaystyle \gamma: [-\pi,\pi ] \to \mathbb{C}$ with $\displaystyle \gamma(t) = Re^{it}$. So the integral will look exactly like before. Just not from $\displaystyle 0$ to $\displaystyle 2\pi$ but instead $\displaystyle -\pi$ to $\displaystyle \pi$. However, this is not an issue here because if you replace $\displaystyle t$ by $\displaystyle t+\pi$ 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 $\displaystyle (-\infty,0]$ and so when you compute the logarithm of $\displaystyle re^{i\theta}$ you need to make sure that $\displaystyle \theta \in (-\pi,\pi]$. 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 $\displaystyle (-\infty,0]$ and so when you compute the logarithm of $\displaystyle re^{i\theta}$ you need to make sure that $\displaystyle \theta \in (-\pi,\pi]$. 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 $\displaystyle (-\infty,0]$. Thanks for the clarification.
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