# Complex integration.

1. Apr 1, 2010

### Urmi Roy

1. The problem statement, all variables and given/known data
I was told to post this kind of question on the homework help section by one of the mentors,even though I'm not sure it is appropiate.
Anyway,I'm doing complex integration now,so I need to get some important concepts cleared.
I'll post my doubts in points...

1.firstly,in complex integration,only curves with nonzero and continuous derivatives are considered....but usually,for integration,all we need,is to take a continous curve,with a derivative defined at each point,not necessarily a non-zero derivative.

2. Whenever I read about complex integration,it's always line integrals....isn't the riemann sum concept of integration applicable to complex functions,where we're simply calculating the area under a curve?

3.Why is simple connectedness a necessary condition for complex integration?
(my book says it's necessary everywhere,without stating any specific reason.)

4.In complex integration,does integration refer to 'integration over a certain curve',or over a certain 'domain area'?

5. It is found that the complex integrals between two fixed points taken over different paths are not always equal...this is the same in real functions isn't it? In that respect,we could perhaps introduce the concept of 'conservative fields as we do in vector calculus' to complex integrals,....but we don't..why? I f we did,we could find out those complex functions that are path independant by using curl=0.

2. Relevant equations

3. The attempt at a solution

2. Apr 8, 2010

### Urmi Roy

It's been a week now...and still no replies :(

3. Apr 8, 2010

### kof9595995

1.Are you sure about your statement? I frequently encountered integrations along a square or rectangular contour.
2. There's no contradiction, line integrals can be defined in terms of riemann sum
3. The reason, i guess, is that the referred complex integrals only specify the limits of integration(for analytic integrands), then the reason is the same as in the vector calculus, you need simple connectedness to ensure the field is conserved. But if the contour is specified, i don't think simple connectedness is really necessary
4. By the way it is defined, it should be 'integration over a certain curve'
5. Sure, you can view the analytic functions as curl=0 vector field, according to cauchy-riemann condition (actually CR condition is even stronger than curl=0 if i'm correct), and in complex analysis analytic functions are what we primarily focus on

4. Apr 9, 2010

### Urmi Roy

It says so in Kreysig,Advanced Engineering Mathematics,page 705....I don't know if I misinterpreted it.

Well...in the first place,ever since I was introduced to the concept of line integrals,I have been thinking that it doesn't exactly fit in to the concept of riemann sum

....on wikipedia,it says that the line integrals and surface integrals are distinct from ordinary intergrals,since they represent weighted sums....inspite of this,can we indeed define line integrals in terms of riemann sum?

It's strange to note,that when we define continuity of a complex function at a particular point,we consider not only the Left hand limits and right hand limits at that point...we approach the point along direction,along different curves,and only when the result is consistent for all these integrals,we say it is continuos at that point.

How do I explain this ? Clearly,here, we don't consider a particular curve...we are considering a region.

5. Apr 9, 2010

### kof9595995

Then you need to offer me some accessible references.

http://en.wikipedia.org/wiki/Line_integral#Definition
You can see the derivation starts from the Riemann sum definition.
Are you still talking about integration? Continuity of course needs to be justified over a small region according to epsilon-delta definition, because the independent variable z can vary on a 2-d plane.
"and only when the result is consistent for all these integrals,we say it is continuos at that point." I really can't figure out what you meant, why define continuity in terms of integrals?? i've never seen it before(I don't even know if it's possible to define it that way)

Last edited: Apr 9, 2010
6. Apr 10, 2010

### Urmi Roy

I'll type out the paragraph in question...
"Complex definite integrals are called(complex) line integrals.
The integrand f(z) is integrated over a given curve C in the complex plane,called the path of integration.
We assume C to be a smooth curve,that is,C has a continuous and non zero derivative dz/dt at each point.Geometrically,this means that C has a unique and continuously turning tangent."
Then,they continue to defiine integrals,and so on.

I think I'll have to look up what caused this confusion in me in the first place...perhaps I've got my mind entangled in various concepts...thanks for pointing this out.I'll be sure to let you know if I face any further confusion in regard to this.

7. Apr 10, 2010

### Urmi Roy

By the way,in Kreysig,they define two methods of integration...the first one is supposed to be only for analytic functions,and the second is for any complex function.....why do we distinguish the process of integration for analytic and non analytic functions?

The first method is said to be the analogue of the formula from calculus.and is said to be an indefinitie integral....however,the author puts limits in the integral,so it;s hard to see why it is indefinite.

The first theorem says,
Let f(z) be analytic in a simply connected domain D.Then there exists an indefinite integral of f(z) in the domain D and for all paths in D joining two points Zo and Z1 in D,we have
(integral within limits Zo to Z1)f(z)dz =F(Z1)-F(zo).

The second theorem is :
Let C be a piecewise smooth path,represented by z=z(t), where t lies between a and b.Let f(z) be a continuous function on C,then

the integral of f(z) over C=

(integral within limits a to b)f(z)dz = (integral within limits a to b)f[z(t)]z'(t)dt

(Sorry,I don;t know how to use latex codes).

8. Apr 10, 2010

### kof9595995

Em.. I think I made a mistake, even for a horizontal path, dz/dt is not zero(z=x(t)+iy(t)), what i had in mind was actually dy/dx. dz/dt=0 requires dx/dt=0 and dy/dt=0 simultaneously.
Then the reason we require dz/dt=dx/dt+idy/dt is nonzero is, i believe, to make sure the path is well-behaved, i.e. we can always bijectively parametrize the path locally,or in other words, in a small enough region there always exists a function y=f(x) or x=f(y), this is guaranteed by the implicit function theorem.
This is also analogous to the line integral in vector calculus, look at the definition here:
http://en.wikipedia.org/wiki/Line_integral#Definition
"where r: [a, b] → C is an arbitrary bijective parametrization of the curve C such that r(a) and r(b) give the endpoints of C."

EDIT: wait a minute, now im not so sure about nonzero dz/dt will guarantee implicit function exist, let me think it through

Last edited: Apr 10, 2010
9. Apr 10, 2010

### kof9595995

By indefinite integral he only means the anti-derivative is well defined for analytic functions, because of the path independence. By putting limits, i think he's just trying to emphasize the path independence(result only depends on the end points)
The second definition is more general, but when the function is analytic, the first and the second will be the same.

10. Apr 10, 2010

### Urmi Roy

How does the curve being 'well behaved' mean that there has to be some y=f(x) or x=f(y)?

Also,for any point,isn't there always some y=f(x) or x=f(y)?It seems to me that even if the derivative at the point is zero,we could have some y=f(x) and x=f(y)...for example y=(x)cubed...it has a zero derivative at 0...but we still have an explicit definition of y=f(x).

11. Apr 10, 2010

### Urmi Roy

Antiderivatives of a function f(z) are just curves that when differentiated give us the required function f(z)..right? Each of the antiderivatives differ by a constant term
....then why did he need to bring in path independance and all that?
Besides,if integration is defined for a function,the antiderivative must be defined....then why did hw have to make a separate theorem for that?

I still can't see why the second is more general...and again,why does a difference arise in intergration of analytic and non-analytic functions.

12. Apr 11, 2010

### kof9595995

No. sometimes you just can't define y=f(x) or x=g(y) locally, one example is a self-crossed curve, you can't define a explicit function where it crossed itself(i don't mean you can't find the expression of f or g, they just don't exist)
see here: http://www.applet-magic.com/implicit.htm
But the thing is, now i don't think nonzero dz/dt would for sure define such a "well-behaved" curve, so i am not sure why it is required

13. Apr 11, 2010

### kof9595995

First let's take the 2nd definition as the fundamental one. Then if the integrand is analytic and the region chosen is simply connected, the integral will be independent with the integration paths, which is exactly the same as in vector calculus, i.e conservative field. Notice in the first definition(actually i'd rather call it a theorem than definition), F(z1)-F(z0) only gives you one result no matter which path you choose, to make sense out of it the integral must be path independent, that's why analytic is required.
In short, take the second as the definition and the first as a theorem

14. Apr 12, 2010

### Urmi Roy

Could I put it like this:

If I start out with the second one,and put f(z) as an analytic function and the region as a simply connected one,then we would automatically arrive at the first one....since the integral is independant of the path..and as you said,he says its and indefinite integral,simply to emphasize that the antiderivatives are well defined at each point....and ofcourse,if we put the antiderivatives within limits,it can easily be made a definite integral.

15. Apr 12, 2010

### Urmi Roy

Thanks,I've understood this.

Thanks for trying,anyway.....Do tell me if you find any other explanation...let's hope someone gives an answer to this for both of us.

16. Apr 12, 2010

### Urmi Roy

Sorry,kof9595995 ,
I posted these questions on another forum,but noone replied...and again,I'm pretty desperate to find out the answers to these....please could you give a try?
Sorry for the botheration....

1. If we have an annulus on a complex plane,is the entire region within the inner circle a set of singularities?

2. An isolated singularity is said to be a point such that there are no other singlularities in its neighbourhood....such a singularity exists only when the principal part(the part of the laurent series with the negative powers) is an infinite series.....please state the reason as to why this criteria exists.

3.The residue at a point is said to be the coefficient of the first negative term in the laurent series...how ?(please give atleast a sketch of the proof).

4. What is the significance of a residue on a complex plane?(i.e How does the function behave differently at a residue?)

5. How are poles different from other singularities?(my book says functions behave differently around poles and other kinds of singularity...why?)

17. Apr 12, 2010

### zpconn

If you are considering a function defined and analytic on an annulus, then in general one would not refer to the points inside the inner disc as singularities. Certainly they are not isolated singularities. The function simply isn't defined there.

That said, there aren't strict rules on how the word singularity is used outside the context of isolated singularities. I would very much advise that you do not use it in this context, though.

This is false. 1/z is its own Laurent series in a neighborhood of 0, and 0 is an isolated singularity--it's a pole.

It's a definition. Definitions aren't proved.

Are you looking for a proof of the residue theorem? That is something you could find in a textbook; it's not suitable for me to reproduce such a proof here.

A residue is just a number associated with an isolated singularity of a meromorphic function. Really there's nothing more to it. It's useful because of the residue theorem.

If f is meromorphic and has a pole at a, then in a deleted neighborhood of a the function g(z) = 1/f(z) is holomorphic. g will have a removable singularity at a, so g has an analytic extension satisfying g(a) = 0.

g does not vanish identically, so a is a zero of finite order, say h. Hence by Taylor's theorem with remainder g(z) = (z - a)^h * j(z) where j(a) != 0. f now has the representation f(z) = (z - a)^(-h) * k(z) where k(z) = 1/j(z) is analytic and nonzero in a neighborhood of a.

18. Apr 13, 2010

### Urmi Roy

Thanks for the clarifications,zpconn

That's good...I wasn not looking for a proof of the residue theorem,since I already have that...actually I thought that the 'residues',with their fancy name and everything must have some very important concept associated with them.....however,I realise now that they have importance due to the fact incidentally fall into the formula to calculate the path integral around a multi-conected region...and don't have any other specific importance,as you said.

I've understood the rest...thankyou once again.

19. Apr 13, 2010

### Urmi Roy

Okay,here are my last few questions..all problem oriented.

1.Are the expansions of (1+x)^(-1), (1+x)^(-2) Taylor expansions?
...but they're also used in problems involving the Laurent series...then are these same for the Taylor series and Laurent series in specific domains?(Uptill today,I thought these expansions were only satndard binomial expansions)

2.Why is |z|<1 such an important criteria whenever we expand a function of z(for both laurent and taylor series)? (whenever we do a sum,we make sure that the z in the denominator has its modulus less than one...else we manipulate it so that the mod does become less than one....but nothing of this sort was specified in the theory for the power series.)

3.expand z(e^(-z)) in Laurent series about 0.....this sum was done simply by using the standard expansion of e^(z),that we even use for real functions)...(putting a minus before the z) in my book ....but laurent series is used only about singularities! The given function doesn't have a singularity at z=0,so how can we use Laurent series for it?

4.What does it imply if a particular laurent series expansion doesn't contain any negative powers?
(e.g (1-cosz)/z doesn't contain any negative powers about z=0)
I'm guessing that in such cases,Laurent expansion=Taylor expansion.

5.If it is not specified,in a sum,and it just says 'expand f(z) about z=(something)', do we take laurent series or taylor series?
(My teacher said that we use only the standard expansions,so it is immaterial if it's the laurent series or taylor series...somehow I didn't like her answer)

6.Expand z/(z-1)(z+2) about -2...do we go for laurent series,since z=-2 is a singularity(in my book,they again use the standard binomial expansion for (1+x)^(-1), so I'm not sure if they've used Taylor or Laurent series)?

20. Apr 16, 2010

### Urmi Roy

Okay,I have my maths exam on Monday...someone atleast tell me why the laurent series is always expanded as a Taylor series.....we don't do anything different for a taylor series and laurent series...like in z/(z-1)(z+2) about -2...(in the book they again use the standard binomial expansion for (1+x)^(-1))