How to solve Cauchy Residue Integral?

[wolly]

[math]\int_Γ z^2e^{2z/(z+1)} dz[/math]
[math]|z-\frac{1}{2}|=R[/math]
[math]R\neq\frac{3}{2}[/math]
[math]R>0[/math]
I know to use the modulus and get the points at x and y and plot them but what I want to find out is how does the radius R descreases and increases by 3/2 as [math]R<3/2[/math] and [math]R>3/2[/math]?
How does this interval appears?
Also in my contest textbook for complex analysis how does the first condition which is smaller by 3/2 makes the integral 0?
I know that the pole is [math]z_0=-1[/math] because [math]z_0+1=0[/math]It's in roumanian but it says that there are 2 cases
If [math]R<3/2[/math]
[math]I=0[/math]
If [math]R>3/2[/math]
[math]I=2*\pi*i*Rez(g,-1)[/math]
When I tried to use the integral I got [math]\int_Γ\frac{2z}{(z+1)^2}[/math].Is it wrong?I used the formula [math]\frac{1}{2*\pi*i}\int_Γ\frac{f(z)}{z-z0}[/math]where [math]f(z)=\frac{2z}{z+1}[/math]
Also can someone explain why






and





This is what I did






Can someone please explain me how to apply the Laurent series in my problem?
I have a few courses in romanian and there are not a lot of romanians here but I'm participating in a student competition of complex analysis(Traian Lalescu)

 

[wolly]

Why îs the first case 0?

 

[logistic_guy]

Hi wolly.

Why îs the first case 0?

Because the singularity is outside the contour.

The idea is simple. You are given a contour. If the singularity lies inside it, use Residue Theorem, if it is outside, then the integral is zero and this is what happened in the case \(\displaystyle R<\frac{3}{2}\).

 

[wolly]

Also how did S come from?
@logistic_guy

Is it based on [math](z+1)^2−2(z+1)+1[/math] and if it is is it Taylor?

 

[blamocur]

I know that the pole is z0=−1z_0=-1z0=−1

So for which values of R is the pole inside [imath]\Gamma[/imath] ?

 

[wolly]

Hi wolly.


Because the singularity is outside the contour.

The idea is simple. You are given a contour. If the singularity lies inside it, use Residue Theorem, if it is outside, then the integral is zero and this is what happened in the case \(\displaystyle R<\frac{3}{2}\).

And how could I calculate that as an integral? Is there a theorem for that because I didn't find it in my textbook!

 

[wolly]

So for which values of R is the pole inside [imath]\Gamma[/imath] ?

Well I know that this is a circle and the R=0,R=9/4

Is this correct?

 

[wolly]

So for which values of R is the pole inside [imath]\Gamma[/imath] ?

It doesn't really show me in my course how the pole is or is not inside gamma.

 

[logistic_guy]

Also how did S come from?
@logistic_guy

Is it based on [math](z+1)^2−2(z+1)+1[/math] and if it is is it Taylor?

The expansion,
\(\displaystyle e^z = \sum_{n=0}^{\infty}\frac{z^n}{n!}\)

Then,

\(\displaystyle e^{\frac{-2}{z+1}} = \sum_{n=0}^{\infty}\frac{(-2)^n}{n!(z+1)^n}\)
This is a Laurent series. This is what it is shown over there.

And how could I calculate that as an integral? Is there a theorem for that because I didn't find it in my textbook!

By Laurent series.

 

[logistic_guy]

Well I know that this is a circle and the R=0,R=-3/2 and R=3/2
View attachment 39341
Is this correct?

Yes correct. Now you can see clearly that \(\displaystyle -1\) lies exactly on the circumference of the circle. But the question says \(\displaystyle R \neq \frac{3}{2}\), so we have to take the two cases, \(\displaystyle R>\frac{3}{2}\) and \(\displaystyle R<\frac{3}{2}\).

\(\displaystyle R < \frac{3}{2}\) does not include the singularity, so you must assume \(\displaystyle R > \frac{3}{2}\) includes it. Otherwise you will get zero for both cases.

It doesn't really show me in my course how the pole is or is not inside gamma.

The singularity lies exactly on the circumference!

 

[logistic_guy]

Here where the calculations of your book went wrong.

The residue at \(\displaystyle z = -1\) is \(\displaystyle c_{-1} = \left(-\frac{2^3}{3!} - \frac{2^3}{2!} - \frac{2}{1!}\right)e^2\)

But your book wrote \(\displaystyle -\frac{2^3}{3!}\) as \(\displaystyle +\frac{2^3}{3!}\) which gives the wrong residue!

 

[blamocur]

It doesn't really show me in my course how the pole is or is not inside gamma.

You've figured out where the pole is. Now you have an equation for a circle which depends on R. If R=1 is the pole inside the circle? How about for R=2?

 

[wolly]

You've figured out where the pole is. Now you have an equation for a circle which depends on R. If R=1 is the pole inside the circle? How about for R=2?

No the pole is not inside the circle!

 

[blamocur]

Reply

No the pole is not inside the circle!

Whether the pole is inside or outside the circle depends on the circle radius.