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 $\int_\gamma f(z) \,dz\,=0$ , where $\gamma$
is the circle with centre 0 and radius 1, oriented as you prefer, and when:

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

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

Do I proceed like:

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

or

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

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?

2. 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 $\int_\gamma f(z) \,dz\,=0$ , where $\gamma$
is the circle with centre 0 and radius 1, oriented as you prefer, and when:

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

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

Do I proceed like:

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

or

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

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.

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

ARgggghhh....

4. 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 $\int_\gamma f(z) \,dz\,=0$ , where $\gamma$
is the circle with centre 0 and radius 1, oriented as you prefer, and when:

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

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

Do I proceed like:

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

or

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

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 $C=\gamma(t),a\leq t \leq b$, then:

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

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

By way of Cauchy-Gousat, $\frac{z}{z-3}$ is analytic outside $z=3$ so that $\mathop\oint\limits_{|z|=1} \frac{z}{z-3}dz=0$

5. 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

6. Great!

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

7. Originally Posted by MathSucker
Great!

Pardon my stupidity, but how do you get $\left(1+\frac{3}{z-3} \right)$ from $\frac{z}{z-3}$ ? 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: