# Complex Line Integrals

• Nov 18th 2008, 03:47 PM
robeuler
Complex Line Integrals
Here I go again...

All of these line! integrals are on the circle |z|=2, with counterclockwise orientation.

1. int(z^104)dz
2. int(z*conjugate(z))d|z|
3. int(conjugate(z))dz
4. int(cos(z)/z^3)dz
5. int(sin(z)/(z^2-((pi)/2)^2)dz

Unfortunately I cannot use the residue theorem.

For numbers 4 and 5 I have been trying to write the integrands as power series, but I am having trouble relating those to line integrals.

For numbers 1, 2 and 3 I feel like I have to split them up into a real and imaginary parts and then do two integrals, but I am having trouble doing this.

And throughout I am always wondering why the line integral is evaluated at a circle of radius 2, instead of the usual 1. I think some of the functions have singularities in that region, but I'm not sure what that does.
• Nov 18th 2008, 06:15 PM
robeuler
Bump. I could really use a hint for any of these.
• Nov 18th 2008, 06:59 PM
TheEmptySet
Quote:

Originally Posted by robeuler
Here I go again...

All of these line! integrals are on the circle |z|=2, with counterclockwise orientation.

1. int(z^104)dz
2. int(z*conjugate(z))d|z|
3. int(conjugate(z))dz
4. int(cos(z)/z^3)dz
5. int(sin(z)/(z^2-((pi)/2)^2)dz

Unfortunately I cannot use the residue theorem.

For numbers 4 and 5 I have been trying to write the integrands as power series, but I am having trouble relating those to line integrals.

For numbers 1, 2 and 3 I feel like I have to split them up into a real and imaginary parts and then do two integrals, but I am having trouble doing this.

And throughout I am always wondering why the line integral is evaluated at a circle of radius 2, instead of the usual 1. I think some of the functions have singularities in that region, but I'm not sure what that does.

You can parameterize the circle $\displaystyle |z|=2 \iff z=2e^{2\pi i t} \implies dz=4\pi i e^{2\pi i t}dt$

So the first one becomes

$\displaystyle \int_{0}^{1}(2e^{2 \pi i t})^{104}(4 \pi i e^{2 \pi i t}dt)$

$\displaystyle 2^{106}\pi i \int_{0}^{1}e^{210 \pi i t}dt=$

$\displaystyle 2^{106}\pi i \left( \frac{1}{210 \pi i }\right)e^{210 \pi i t} \bigg |_{0}^{1}=0$

This should help with the rest good luck (Rock)
• Nov 18th 2008, 07:15 PM
robeuler
Quote:

Originally Posted by TheEmptySet
You can parameterize the circle $\displaystyle |z|=2 \iff z=2e^{2\pi i t} \implies dz=4\pi i e^{2\pi i t}dt$

So the first one becomes

$\displaystyle \int_{0}^{1}(2e^{2 \pi i t})^{104}(4 \pi i e^{2 \pi i t}dt)$

$\displaystyle 2^{106}\pi i \int_{0}^{1}e^{210 \pi i t}dt=$

$\displaystyle 2^{106}\pi i \left( \frac{1}{210 \pi i }\right)e^{210 \pi i t} \bigg |_{0}^{1}=0$

This should help with the rest good luck (Rock)

Will the parametrization of z conjugate just be negative the parametrization of z? Also: how do evaluate the absolute value of dz? Does the dt matter?

Thank you so much for your help.