Complex integration

• September 27th 2010, 04:16 PM
Danneedshelp
Complex integration
Q: Show that $\oint_{C}\frac{1}{z-z_{0}}dz$ is equal to $2\pi\\i$ for every circle $C$ centered at $z_{0}$.

First, I wrote

$\oint_{C}\frac{1}{z-z_{0}}dz=\oint_{C}\frac{1}{z-z_{0}}\frac{\bar{z-z_{0}}}{\bar{(z-z_{0})}}dz=\oint_{C}\frac{(\bar{z}-\bar{z_{0}})}{|z-z_{0}|^{2}}dz$

Now,

$\bar{z}-\bar{z_{0}}=x-iy-x_{0}+iy_{0}=(x-x_{0})-i(y-y_{0})$.

So,

$(x-x_{0})-i(y-y_{0})(dx+idy)=(x-x_{0})dx-i(y-y_{0})dx+i(x-x_{0})dy+(y-y_{0})dy$.

Thus,

$\frac{1}{|z-z_{0}|^{2}}[\oint_{C}(x-x_{0})dx+(y-y_{0})dy+i(\oint_{C}-(y-y_{0})dx+(x-x_{0})dy)]$

Here is where I am stuck. Firstly, I am assuming I will need to use the parameterization $x=cos(t)$, $y=sin(t)$, with $0\leq\\t\leq\\2\pi$.

So, I want to integrate along $C=\{(x,y):(x-x_{0})^{2}+(y-y_{0})^{2}=r\}$. I am not sure how to use the parameters and integrate along this curve.

Any help would be appreciated.

Thanks
• September 27th 2010, 05:14 PM
Ackbeet
I don't think your general method is correct. I would parametrize the circle like this:

Let $z=z_{0}+r\,e^{i\theta},$ for $r$ constant and $\theta\in[0,2\pi].$ Thus, $dz=i\,r\,e^{i\theta}\,d\theta.$ What does your integral become?
• September 28th 2010, 12:44 PM
Danneedshelp
Quote:

Originally Posted by Ackbeet
I don't think your general method is correct. I would parametrize the circle like this:

Let $z=z_{0}+r\,e^{i\theta},$ for $r$ constant and $\theta\in[0,2\pi].$ Thus, $dz=i\,r\,e^{i\theta}\,d\theta.$ What does your integral become?

Thanks.

Even so, we have not covered that material and his hint was to use greens thereom. We have just been doing integrals by seperating the real and imagenary parts into two seperate integrals, the one with the imaginary integrand has a constant i out front.
• September 28th 2010, 01:23 PM
Ackbeet
You can do Green's Theorem. So, in reviewing your work, I'd say you're good up to here:

$\displaystyle{\frac{1}{|z-z_{0}|^{2}}\left[\oint_{C}[(x-x_{0})dx+(y-y_{0})dy]+i\oint_{C}[-(y-y_{0})dx+(x-x_{0})dy)]\right]}.$

Now, what you want to do is apply Green's Theorem on each integral. Green's theorem is going to reduce your integrals quite a bit. Recall that Green's Theorem states that

$\displaystyle{\oint_{C}(L\,dx+M\,dy)=\iint_{D}\lef t(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right)dx\,dy}.$

Once you have constructed your double integrals, I'd recommend converting to polar coordinates. The result pops out fairly readily.
• September 28th 2010, 04:33 PM
Danneedshelp
Quote:

Originally Posted by Ackbeet
You can do Green's Theorem. So, in reviewing your work, I'd say you're good up to here:

$\displaystyle{\frac{1}{|z-z_{0}|^{2}}\left[\oint_{C}[(x-x_{0})dx+(y-y_{0})dy]+i\oint_{C}[-(y-y_{0})dx+(x-x_{0})dy)]\right]}.$

Now, what you want to do is apply Green's Theorem on each integral. Green's theorem is going to reduce your integrals quite a bit. Recall that Green's Theorem states that

$\displaystyle{\oint_{C}(L\,dx+M\,dy)=\iint_{D}\lef t(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right)dx\,dy}.$

Once you have constructed your double integrals, I'd recommend converting to polar coordinates. The result pops out fairly readily.

Ahhhh, I see. The first integral will go to zero and I will end up with the integrand 2 after taking the partails of the other chunk and applying greens thereom. So, if convert to polar and integrate over the region $D=\{(r,\theta):0 I end up with I am looking for after distributing the $\frac{1}{|z-z_{0}|^{2}}$ and the $i$.

Thanks for the help, I appreciate it. I also carried out the method you mentioned in your first post and got the answer that way.

Thanks again.
• September 29th 2010, 05:47 AM
Ackbeet
Yep, that's exactly right. You're welcome, and have a good one!