• Mar 21st 2010, 09:57 AM
MathSucker
I am trying to work through these simple exercises but it invariably blends together and I get lost. Here is one of them:

Use the Complex FTC and/or Cauchy-Goursat Theorem to show that $\displaystyle \int_\gamma f(z) \,dz\,=0$ , where $\displaystyle \gamma$
is the circle with centre 0 and radius 1, oriented as you prefer, and when:

1. $\displaystyle f(z)=\frac{z}{z-3}$

I'm not sure how to proceed. I know that $\displaystyle \gamma \, : \, e^{it}$ , $\displaystyle t \in [0,2\pi)$

Do I proceed like:

$\displaystyle \int_0^{2\pi} \frac {e^{it}}{e^{it}-3} \, dt \,$

or

$\displaystyle \int_0^{2\pi} \frac {cos(t)+isin(t)}{cos(t)+isin(t)-3} \, dt \,$

Do I just integrate the original equation and forget about polar coordinates?

Surely this is a simple question. Should I be doing something else? Is there any way to avoid tedious integration by parts?
• Mar 21st 2010, 10:08 AM
chiph588@
Quote:

Originally Posted by MathSucker
I am trying to work through these simple exercises but it invariably blends together and I get lost. Here is one of them:

Use the Complex FTC and/or Cauchy-Goursat Theorem to show that $\displaystyle \int_\gamma f(z) \,dz\,=0$ , where $\displaystyle \gamma$
is the circle with centre 0 and radius 1, oriented as you prefer, and when:

1. $\displaystyle f(z)=\frac{z}{z-3}$

I'm not sure how to proceed. I know that $\displaystyle \gamma \, : \, e^{it}$ , $\displaystyle t \in [0,2\pi)$

Do I proceed like:

$\displaystyle \int_0^{2\pi} \frac {e^{it}}{e^{it}-3} \, dt \,$

or

$\displaystyle \int_0^{2\pi} \frac {cos(t)+isin(t)}{cos(t)+isin(t)-3} \, dt \,$

Do I just integrate the original equation and forget about polar coordinates?

Surely this is a simple question. Should I be doing something else? Is there any way to avoid tedious integration by parts?

Simply apply the residue theorem.
• Mar 21st 2010, 12:05 PM
MathSucker
I think I'll be getting on to that next, but for now I have to use the complex FTC or Cauchy-Goursat...

ARgggghhh....
• Mar 21st 2010, 12:15 PM
shawsend
Quote:

Originally Posted by MathSucker
I am trying to work through these simple exercises but it invariably blends together and I get lost. Here is one of them:

Use the Complex FTC and/or Cauchy-Goursat Theorem to show that $\displaystyle \int_\gamma f(z) \,dz\,=0$ , where $\displaystyle \gamma$
is the circle with centre 0 and radius 1, oriented as you prefer, and when:

1. $\displaystyle f(z)=\frac{z}{z-3}$

I'm not sure how to proceed. I know that $\displaystyle \gamma \, : \, e^{it}$ , $\displaystyle t \in [0,2\pi)$

Do I proceed like:

$\displaystyle \int_0^{2\pi} \frac {e^{it}}{e^{it}-3} \, dt \,$

or

$\displaystyle \int_0^{2\pi} \frac {cos(t)+isin(t)}{cos(t)+isin(t)-3} \, dt \,$

Do I just integrate the original equation and forget about polar coordinates?

Surely this is a simple question. Should I be doing something else? Is there any way to avoid tedious integration by parts?

The CFTC states that if the antiderivative is defined in a domain containing the contour $\displaystyle C=\gamma(t),a\leq t \leq b$, then:

$\displaystyle \oint f'(z)dz=f(\gamma(b))-f(\gamma(a))$

and so if it's a closed contour and the antiderivative is analytic over the entire contour, then the integral is zero.
Then $\displaystyle \mathop\oint\limits_{|z|=1} \frac{z}{z-3}dz=\mathop\oint\limits_{|z|=1} \left(1+\frac{3}{z-3}\right)dz=(z+3\log(z-3))\biggr|_{z=1}^{z=1}=0$ since $\displaystyle \log(z-3)$ is analytic over the unit circle and in fact in any domain not including the point $\displaystyle z=3$.

By way of Cauchy-Gousat, $\displaystyle \frac{z}{z-3}$ is analytic outside $\displaystyle z=3$ so that $\displaystyle \mathop\oint\limits_{|z|=1} \frac{z}{z-3}dz=0$
• Mar 21st 2010, 12:17 PM
chiph588@
Quote:

Originally Posted by MathSucker
I think I'll be getting on to that next, but for now I have to use the complex FTC or Cauchy-Goursat...

ARgggghhh....

Well this problem is a direct result of Cauchy-Goursat.

Cauchy's integral theorem - Wikipedia, the free encyclopedia
• Mar 21st 2010, 12:29 PM
MathSucker
Great!

Pardon my stupidity, but how do you get $\displaystyle \left(1+\frac{3}{z-3} \right)$ from $\displaystyle \frac{z}{z-3}$ ? My algebra is a little off.
• Mar 21st 2010, 12:39 PM
shawsend
Quote:

Originally Posted by MathSucker
Great!

Pardon my stupidity, but how do you get $\displaystyle \left(1+\frac{3}{z-3} \right)$ from $\displaystyle \frac{z}{z-3}$ ? My algebra is a little off.

That's just long division. Try and get good at it as it's useful in Complex Analysis for example when computing residues of certain functions: