# Homework Help: Help with Cauchy Integral Formula

1. Oct 29, 2013

Use Cauchy Integral Formula to solve:

∫ [(5z² - 3z + 2)/(z-1)³] dz

C is any closed simple curve involving z=1. (z is a complex)

Thanks and sorry for my poor english, it's not my first language.

2. Oct 29, 2013

### Dick

State the Cauchy Integral Formula. Then express your function in a simpler form using partial fractions. Do something.

Last edited: Oct 29, 2013
3. Oct 30, 2013

Ok, I put it in partial fractions:

[(5z² - 3z + 2)/(z-1)³] = 5/(z-1) + 7/(z-1)² + 4/(z-1)³

But I have no idea how to solve it using Cauchy... can anybody at least give me a hint? Or try to explain something, a next step?
I've seen a few videos and examples around internet but when I try to solve it, I just can't apply... I really need help on this.

4. Oct 30, 2013

### HallsofIvy

You were asked to "state the Cauchy Integral Formula". Are you saying you do not know what it is? Surely if your textbook refers to it, it must be stated in that chapter.

(You really don't need the full "integral formula". It is sufficient to know that the integral around two different paths is the same as long as there are no singularities in between the two paths. Here you can take the integral around a circle centered at 0, with radius 1.)

5. Oct 30, 2013

Like, with the partial fractions I will have now

i2π(f(5)) + (f(7)) + (f(4))

where the f(a) = (5z² - 3z + 2).

Is this that I have to do?

6. Oct 30, 2013

### vanhees71

State the integral formula! Then you should know, what's to do. Of course you must not take any curve going through the singularity at $z=1$ as suggested by HallsofIvy, but any closed curve with this point in the interior of the area enclosed by it!

7. Oct 30, 2013

Sorry, I know I'm really dumb ...but anyways

I have this Cauchy Int. Formula : ∫F(z)/((z-z0))n

My doubt is about this countour, it just says it's a simple closed one with z =1.In this case, do I have a circle(for exemple) centred in (0,0) with r=1?

Then my z0=1
and my n= 3
and my f(z)= 5z²-3z+2

Then I have that Cauchy In. Formula = [2πi /(3)!] f(3)(z0)

And f(3) means that I have to derivate f(z) 3 time, is it?

Last edited: Oct 30, 2013
8. Oct 30, 2013

### Dick

You want a circle around the point z=1. That's the point (1,0) if you are looking at x,y coordinates in the complex plane. And the radius or shape doesn't really matter. It just has to wind around z=1. Apply the Cauchy Integral Formula to each term in your partial fractions form. Or if you can do it all at once like you are doing, but read the Formula statement more carefully if you think you differentiate three times. You don't.

Last edited: Oct 30, 2013
9. Oct 30, 2013

Here's the formula I know for cauchy's integral

Where n is the radius, and f3(z) means I have to derivate F(z) 3times. At least it's what's on the only exercise we had in class...(f(z) was an exponencial at tha case)

Decomposing it in partial fraction doesn't help. If was a function type of 1/((z-a)(z-b)) or a quadratic function I would know...then I would appy another version of this formula.

But 1/(z-1)³ gives me fractions with power on the denominator, it doesn't help.

I have this specific function(that I posted) when you have powers on the denominator, but it demands de radius...

I'm stucked, someone explain how I must do it?!

#### Attached Files:

• ###### cauchy general.png
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1.3 KB
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10. Oct 30, 2013

### Dick

I can barely read the formula you've attached but it looks like you have $(z-z_0)^{n+1}$ in the denominator. So if you have $(z-1)^3$ then n+1=3 and you want to use n=2 in your formula. Not n=3. Why are you saying n is the radius? I don't see any radius in that formula.

11. Oct 30, 2013

If you click with the right button and open it in a new tab you'll see it clearly. Or click in the image, then will appear a black image when you can barely see it, just click in this black image and it will open a new tab with the formula.

n is the |z|, so n is the radius. In the example in class, that's the information given...

In the problem I wrote C is any closed simple curve involving z=1. Then I assumed |z| = 1, where z = x+ iy
so x² + y² = 1²

I thought it wrong?

12. Oct 30, 2013

### Dick

Thanks, that link works better. Your information is wrong. n isn't the radius. n+1 is supposed to be the power in the denominator. You don't need to know the radius. You just need to know the curve circles z=1. And x^2+y^2=1 doesn't do that. It goes through z=1. A better curve is (x-1)^2+y^2=1, that's actually a circle around z=1. z=1 is the center.

13. Oct 30, 2013

So n+1=3, n=2

And according to that formula I have to do f''(z0)

Where my F(z) = 5z²-3z+2
F''(z)=10

F''(z)[(2∏i)/2!] = ∫5z²-3z+2 / (x-1)³ = 10i∏

Is it?

14. Oct 30, 2013

### Dick

Now you've got it.

15. Oct 30, 2013