Complex Contour Integrals

**JoAdams5000** · October 1st 2010, 07:58 PM

Hello,

I have been working on complex integration in my own time and have managed to solve a few, but these three seriously have me stumped. I was wondering if anyone would be willing to help me understand these problems.

a) $\int_{\gamma } \frac{z}{(z+2)(z-1)}dz$ where $\gamma$ is the circle of radius 4 about the origin, traversed twice in the clockwise direction.

b) $\int_{\gamma } \frac{2z^{2}-z+1}{(z-1)^2(z+1)}dz$ where $\gamma$ proceeds around the boundary of the figure eight formed by two circles of radius 1 with centres 1 and −1 by starting at 0, going once counterclockwise around the right circle followed by going once counterclockwise around the left circle.

c) $\int_{\gamma } \frac{5z-2}{z(z-1)(z-3)}dz$ where $\gamma$ is the circle of radius 2 about the origin.

I thank you all in advance! Your help is very much appreciated.

**Defunkt** · October 2nd 2010, 09:45 AM

What tools are you allowed to use? Cauchy's integral theorem? Residue theorem?

**JoAdams5000** · October 2nd 2010, 11:10 AM

Yes I've read about those two. I'm just learning this on my own so all tools are helpful. Would those two be helpful in the given scenarios?
Thanks

**JoAdams5000** · October 3rd 2010, 06:55 PM

I was also wondering if trying to parameterize the curve would be an option as well, but I have a feeling that trying to solve using the basic definition will not produce the most efficient result. Any ideas?

**Defunkt** · October 4th 2010, 03:22 AM

Hey, sorry for the delay.

For your first post - yes they will be very useful here.

What is absolutely essential here is that you draw the curves that you are integrating over, so that you know how to parameterize them. This is important not because you will use the parametrizations explicitly, but it will give you an idea of how to use the needed theorems.

I'll do the first one for you - try to do the others in a similar way.

Let $\displaystyle I = \int_{\gamma } \frac{z}{(z+2)(z-1)}dz$ where $\gamma$ is as mentioned.
Since this is not a simple curve (it intersects itself) you will have to reparameterize it as a concatenation of two simple curves $\gamma = \gamma_1 * \gamma_1$ where $\gamma_1$ traverses $|z|=4$ once, clock-wise.

By properties of concatenation of curves, you have $\displaystyle I = \int_{\gamma } \frac{z}{(z+2)(z-1)}dz = \int_{\gamma_1} \frac{z}{(z+2)(z-1)}dz + \int_{\gamma_1} \frac{z}{(z+2)(z-1)}dz = 2 \cdot \int_{\gamma_1} \frac{z}{(z+2)(z-1)}dz$

Now, using the residue theorem:
$\displaystyle \int_{\gamma_1} \frac{z}{(z+2)(z-1)}dz = 2 \pi i \sum_{k} Res(f(z), z_k)$ where $\displaystyle f(z) = \frac{z}{(z+2)(z-1)} = \frac{z}{z^2 + z -2}$ and $z_k$ are the poles of $f(z)$ inside $\gamma_1$ - $z_1 = -2, \ z_2 = 1$

Now calculate the residues:
$\displaystyle Res(f(z), -2) = \frac{z}{2z + 1} |_{z = -2} = \frac{-2}{-3} = \frac{2}{3}$
$\displaystyle Res(f(z), 1) = \frac{z}{2z + 1} |_{z = 1} = \frac{1}{3}$

And so $\displaystyle \int_{\gamma_1} \frac{z}{(z+2)(z-1)}dz = 2 \pi i \left( \frac{1}{3} + \frac{2}{3} \right) = 2 \pi i$
And finally $\displaystyle I = 2 \cdot \int_{\gamma_1} \frac{z}{(z+2)(z-1)}dz = 4 \pi i$

If you are not familiar with all of this then you are probably expected to use Cauchy's integral theorem. If you need that I'll also show you how to solve this integral with that.

**JoAdams5000** · October 14th 2010, 07:19 PM

Thanks!

I was wondering if you don't mind could you also demonstrate how to do this using Cauchy's Integral theorem? I appreciate it!

**ComplexXavier** · October 16th 2010, 09:50 PM

I tried solving this problem using CIT and I got these answers.

a) -4ipi
b) 4ipi
c) -4ipi

Has anyone else tried this?

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