# Cauchy.riemann integral theorem or formula

1. Jun 2, 2014

### MissP.25_5

Hello.
How do I start this question? Do I use Cauchy.riemann integral theorem? or Cauchy.riemann integral formula?

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2. Jun 2, 2014

### lurflurf

There are many ways to find the integrals. The key point is if 0 is inside the given circles of outside.
What is the value in each case (0 is inside and 0 is outside)?
For C1 and C2 is 0 inside or outside?

3. Jun 2, 2014

### MissP.25_5

The first one : 0 is outside the circle while the second one is inside. Am I right? Why do we have to know this?

4. Jun 2, 2014

### lurflurf

^You do not have to know, but it helps. Do the integral for any circle centered at the origin. What is the value? That will be the same value as any closed contour that has 0 inside it.

1/z has a pole at z=0

Integrating around it will not be zero.

I am not sure how far into your class you are. That might be a new fact.

5. Jun 2, 2014

### MissP.25_5

Yes, I have learned that 1/z has a pole at z=0 but I am not sure what it means. Do you have a diagram so I can see it?

6. Jun 3, 2014

### MissP.25_5

So, the for no.1, 1/z has a pole at z=0 because the circle is outside of 0? What about the second one? 0 is inside the circle, does that mean the pole is not at z=0?

Last edited: Jun 3, 2014
7. Jun 3, 2014

### MissP.25_5

Ok, I just re-read again about Cauchy's integral theorem and this is what I got. I hope I have understood it right. Please check my solution for each question.

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8. Jun 3, 2014

### lurflurf

^That looks good. For (ii) you should mention that you have changed the path without crossing z=0.

The pole (for 1/z) is always at z=0. The important thing is to know if z=0 is inside the circle or not. As you have said it is not for (i) and it is for (ii). Again I do not know what theorems and methods you have covered so far. You can calculate the two integrals by parametrization for example. The integration path can be moved without changing the value of the integral so long as the path does not cross any points where the function is not differentiable such as z=0 for 1/z. Thus the integral around a circle will have one value if z=0 is inside the circle and another value if z=0 is outside the circle.

9. Jun 3, 2014

### MissP.25_5

How have I changed the integral path? You mean because I integrate it from 0 to 2pi?