Calculate the following contour integrals sing suitable parametr

[bugatti79]

Calculate the following contour integrals using suitable parameterisations

Homework Statement

1)$\oint \frac{1}{z-z_0} dz$ where C is the circle $z_0$ and radius r>0 oriented CCW and $k\ge0$

2) $\int_c |z|^2 dz$ where C is the straight line from 1+i to -1

3. Relevant equations

$\oint f(z)dz=\int_a^b f(z(t))*z'(t)dt$

The Attempt at a Solution

1) let z(t)=e^it for $0\le t \le1$ and $z_0=0$ since it is centred at $(0,0)$

We have $\displaystyle \int_c \frac{1}{z^k} dz=\int_0^{2\pi} \frac{1}{e^{ikt}} i e^{it} dt=\frac{i}{(1-k)}e^{it}|_0^{2\pi}$..?


2) $\int_c |z|^2 dz$ where C is the straight line from 1+i to -1

let $z(t)=(1-t)(1+i)+t(0)=1-t+i(1-t)$ for $0 \le t \le 1$

$z'(t)=-1-i$

$|z(t)|^2=\sqrt((1-t)^2+(1-t)^2)=2t^2-4t+2$

Therefore $\int_c |z|^2 dz=\int_0^1 (2t^2-4t+2)(-1-i)dt$...?

How am I doing so far on both?

Thanks

 

[scurty]

Homework Statement

1)$\oint \frac{1}{z-z_0} dz$ where C is the circle $z_0$ and radius r>0 oriented CCW and $k\ge0$1) let z(t)=e^it for $0\le t \le1$ and $z_0=0$ since it is centred at $(0,0)$

We have $\displaystyle \int_c \frac{1}{z^k} dz=\int_0^{2\pi} \frac{1}{e^{ikt}} i e^{it} dt=\frac{i}{(1-k)}e^{it}|_0^{2\pi}$..?

Your reasoning isn't consistent here. What is k in the original problem? What is $z_0$? Is the circle centered at $z_0$ with positive radius r? I can't help you until that is cleared up.

2) $\int_c |z|^2 dz$ where C is the straight line from 1+i to -12) $\int_c |z|^2 dz$ where C is the straight line from 1+i to -1

let $z(t)=(1-t)(1+i)+t(0)=1-t+i(1-t)$ for $0 \le t \le 1$

$z'(t)=-1-i$

$|z(t)|^2=\sqrt((1-t)^2+(1-t)^2)=2t^2-4t+2$

Therefore $\int_c |z|^2 dz=\int_0^1 (2t^2-4t+2)(-1-i)dt$...?

How am I doing so far on both?

Thanks

Your formula for z(t) is wrong. z(1) equals 0, not -1. You meant to write $z(t) = (1-t)(1+i) + t(-1)$

Your method seems correct after that step. Also, $|z|^2 = z \bar{z}$, it miught be easier to do that, might not.

 

[bugatti79]

Homework Statement

1)$\oint \frac{1}{z-z_0} dz$ where C is the circle $z_0$ and radius r>0 oriented CCW and $k\ge0$

2) $\int_c |z|^2 dz$ where C is the straight line from 1+i to -1

3. Relevant equations

$\oint f(z)dz=\int_a^b f(z(t))*z'(t)dt$

The Attempt at a Solution

1) let z(t)=e^it for $0\le t \le1$ and $z_0=0$ since it is centred at $(0,0)$

We have $\displaystyle \int_c \frac{1}{z^k} dz=\int_0^{2\pi} \frac{1}{e^{ikt}} i e^{it} dt=\frac{i}{(1-k)}e^{it}|_0^{2\pi}$..?


2) $\int_c |z|^2 dz$ where C is the straight line from 1+i to -1

let $z(t)=(1-t)(1+i)+t(0)=1-t+i(1-t)$ for $0 \le t \le 1$

$z'(t)=-1-i$

$|z(t)|^2=\sqrt((1-t)^2+(1-t)^2)=2t^2-4t+2$

Therefore $\int_c |z|^2 dz=\int_0^1 (2t^2-4t+2)(-1-i)dt$...?

How am I doing so far on both?

Thanks

Your reasoning isn't consistent here. What is k in the original problem? What is $z_0$? Is the circle centered at $z_0$ with positive radius r? I can't help you until that is cleared up.

C is the circle with centre z_0 and radius r>0 orientated CCW. I left out the k, it should read

$\oint \frac{1}{(z-z_0)^k} dz$

I don't know how to handle the z_0...?

Your formula for z(t) is wrong. z(1) equals 0, not -1. You meant to write $z(t) = (1-t)(1+i) + t(-1)$

Your method seems correct after that step. Also, $|z|^2 = z \bar{z}$, it miught be easier to do that, might not.

Sorry, that's a mistake on my part. Will review..

 

[scurty]

C is the circle with centre z_0 and radius r>0 orientated CCW. I left out the k, it should read

$\oint \frac{1}{(z-z_0)^k} dz$

I don't know how to handle the z_0...?

Okay! So what is your equation equation for a circle centered at $z_0$ with positive radius? Then use the contour integral formula. You should come out with an answer very similar to what your had in your first post.

$\displaystyle \int_c \frac{1}{(z - z_0)^k} dz=\int_0^{2\pi} \frac{1}{(z_0 + re^{it} - z_0)^k} (ir e^{it}) dt= \dots$

Make sure your anti-derivative is correct, your anti-derivative in your first post was wrong.

Also, is k any real number greater than or equal to zero or any integer greater than or equal to zero? I'm inclined to believe the latter but the choice of what k is will determine what the final answer is.

 

[bugatti79]

Okay! So what is your equation equation for a circle centered at $z_0$ with positive radius? Then use the contour integral formula. You should come out with an answer very similar to what your had in your first post.

$\displaystyle \int_c \frac{1}{(z - z_0)^k} dz=\int_0^{2\pi} \frac{1}{(z_0 + re^{it} - z_0)^k} (ir e^{it}) dt= \dots$

Make sure your anti-derivative is correct, your anti-derivative in your first post was wrong.

Also, is k any real number greater than or equal to zero or any integer greater than or equal to zero? I'm inclined to believe the latter but the choice of what k is will determine what the final answer is.

Yes it is the latter. I see you have taken $z=re^{it}+z_0$ SO the integral should work out to be

$\displaystyle \frac{ir^2}{(1-k)}e^{it(1-k)} |_0^{2\pi}$...?