# Seemingly easy complex line integral

1. Dec 10, 2011

### Jamin2112

1. The problem statement, all variables and given/known data

Find ∫(ez+cos(z))/z dz integrated over C1(0)

2. Relevant equations

Theorem 6.10 (Cauchy's integral formula)

Let f be analytic in the simply connected domain D and let C be a simple closed positively oriented contour that lies in D. If z0 is a point that lies interior to C, then

f(z0) = 1/2πi ∫f(z)/(z-z0) dz

3. The attempt at a solution

So the answer is 4πi, which is of course what you obtain if you invoke Cauchy's integral formula. But our function isn't analytic inside the region over which we're integrating. (???)

2. Dec 10, 2011

### Dick

The theorem doesn't say that f(z)/z is analytic inside the contour. If it were the answer would be 0. It's says that f(z) is analytic inside the contour.

3. Dec 10, 2011

### Jamin2112

http://pinkie.ponychan.net/chan/files/src/132330639382.gif [Broken]

Dang it. My reading comprehension skills suck.

Last edited by a moderator: May 5, 2017
4. Dec 10, 2011

### Jamin2112

I'll ask another related question just to make this thread worthwhile.

5. Dec 10, 2011

### Jamin2112

Section 7.3, Exercise 9.

Find two Laurent series for z-1(4-z)-2 involving powers of z and state where they are valid.

Solution so far.

Obviously f(z)=1/(z(4-z)2) has singularities at z = 0 and z = 4. It is therefore analytic in the two annuli A(0,0,4) and A(0,4,∞). In the first case, choose p = 3, since 0 < 3 < 4. By Laurent's theorem, our c-n will be equal to

1/2πi ∫f(z)/(z-0)-n+1 dz = 1/2πi ∫dz/(4-z)2z-n+2
integrated on the curve C3(0).

But this seems too difficult. There must be a better way. ¿Qué me recomienda Ud.?

6. Dec 10, 2011

### Dick

You generally find Laurent series by doing things like partial fractions expansions and power series expansions, not by evaluating contour integrals. That would be too difficult. Don't you have any examples showing you these sort of techniques?

7. Dec 10, 2011

### Jamin2112

Yeah, brah. One example is finding the Lauret series for f(z) = 3/(2+z-z2).

f(z) = 3/((1+z)(2-z)) = (by partial fraction decomp.) 1/(1+z) + (1/2)*(1/(1-z/2)). I understand how they found out the power series for (1/2)*(1/(1-z/2)) but not for 1/(1+z). (They just refer me to "Corollary 4.2", which, as far as I can tell, doesn't reveal anything about 1/(1+z))

8. Dec 10, 2011

### Dick

If you know how to power expand 1/(1-z/2), I'm not sure why you would have problem expanding 1/(1+z). Write it as 1/(1-(-z)). Now it looks almost the same as the first.

9. Dec 10, 2011

### Jamin2112

Another quick question:

Correct me if I wrong.

f(z) is analytic in a region D
---> f(z) is continuous in D (converse doesn't necessarily hold),
can be represented as a power series convergent in D,
is differentiable on D, and
is integrable on D.

10. Dec 10, 2011

### Jamin2112

11. Dec 10, 2011

### Dick

You aren't just trying to use up your funny graphic things are you? Of course, analytic in D implies differentiable in D. That's what analytic means. And no, it can't necessarily be represented as a single power series valid everywhere in D. Laurent series give you plenty of examples of that. And depends on what you mean by 'integrable'.

12. Dec 10, 2011

### Jamin2112

(I might use an appropriate graphic if I get the chance)

Says Wikipedia:

"In complex analysis, a branch of mathematics, a complex-valued function ƒ of a complex variable z [...] is said to be analytic at a if in some open disk centered at a it can be expanded as a convergent power series"

http://en.wikipedia.org/wiki/Holomorphic_functions_are_analytic

By "integrable" I mean you could integrate it over a curve lying in D. I remember the professor saying something like, "if it's differentiable, it's integrable; and vice versa."

13. Dec 10, 2011

### Dick

If it has a convergent power series around a point, then sure it's differentiable and continuous at that point. And also sure, the integral along a curve is well defined. It might be divergent if the curve is unbounded in some way, like having an open endpoint or going to infinity.

14. Dec 10, 2011

### Jamin2112

Understood, brah. Thanks a million. That's all the questions I have for now; I might hit this place up later tonight, after I work on some practice problems.