# Complex line integral

• Sep 1st 2009, 12:14 PM
arbolis
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?
• Sep 1st 2009, 06:18 PM
ThePerfectHacker
Quote:

Originally Posted by arbolis
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.
• Sep 1st 2009, 06:23 PM
arbolis
Quote:

Originally Posted by ThePerfectHacker
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?
• Sep 1st 2009, 06:26 PM
ThePerfectHacker
Quote:

Originally Posted by arbolis
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.
• Sep 1st 2009, 06:32 PM
arbolis
Quote:

Originally Posted by ThePerfectHacker
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.