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Homework Help: Complex integration.

  1. Apr 1, 2010 #1
    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. jcsd
  3. Apr 8, 2010 #2
    It's been a week now...and still no replies :(
     
  4. Apr 8, 2010 #3
    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
     
  5. Apr 9, 2010 #4
    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.

    Thanks for replying,kof9595995.
     
  6. Apr 9, 2010 #5
    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
  7. Apr 10, 2010 #6
    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.
     
  8. Apr 10, 2010 #7
    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).
     
  9. Apr 10, 2010 #8
    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
  10. Apr 10, 2010 #9
    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.
     
  11. Apr 10, 2010 #10
    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).
     
  12. Apr 10, 2010 #11
    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.
     
  13. Apr 11, 2010 #12
    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
     
  14. Apr 11, 2010 #13
    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
     
  15. Apr 12, 2010 #14
    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.
     
  16. Apr 12, 2010 #15
    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.
     
  17. Apr 12, 2010 #16
    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?)
     
  18. Apr 12, 2010 #17
    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.
     
  19. Apr 13, 2010 #18
    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.
     
  20. Apr 13, 2010 #19
    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)?
     
  21. Apr 16, 2010 #20
    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))
     
  22. Apr 16, 2010 #21
    Well, it is basically an expansion about infinity. When z is very large, 1/z is small and then you can expand in powers of that small parameter.
     
  23. Apr 16, 2010 #22
    How is z very large?....it's specified as -2 here...also,since the expansion is to be done about a singularity....how can the two series be same here(laurent series becomes same as the Taylor series only when there is no singularity)?
     
    Last edited: Apr 16, 2010
  24. Apr 16, 2010 #23
    There are two Laurent expansions around z = -2. In one case you only have one singular term in the Laurent series. In that case you can intepret the singular terms of the Laurent expansion as representing a part fo the partial fraction expansion, in this case involving the 1/(z+2) term. You can then say that if you subtract these terms from the rational function, you get something that is regular at z = 2 and can therefore be expanded in a Taylor series. So, after you do that, you can add back the singular terms to get the expansion of the full function.

    Now, in practice it is actually convenient to reverse this logic and find partial fraction expansions by expanding around the singularities. You then factor out the singular part of the function and expand the rest (which is regular).

    In this case we can write:

    z/[(z-1)(z+2)] = 1/(z+2) z/(z-1)

    The factor z/(z-1) can be expanded around z = -2 in a Taylor expansion. Clearly multiplying that Taylor expansion by 1/(z+2) will give you the Laurent expansion. The singular part of the Laurent expansion thus consists of the single term

    2/3 1/(z+2)

    There is another singularity at z = 1. To obtain the Laurent expansion around that point, you can write:


    z/[(z-1)(z+2)] = 1/(z-1) z/(z+2)

    Expanding the factor z/(z+2) in a Taylor expansion around z = 1 will thus give you the Laurent expansion around z = 1. The singular term of that expansion is thus:

    1/3 1/(z-1)


    Now witness some magic. Let's take our function

    z/[(z-1)(z+2)]

    and subtract from it the two singular parts from those two expansions around the two points z = -2 and z = 1. We then get the function:

    z/[(z-1)(z+2)] - 2/3 1/(z+2) - 1/3 1/(z-1)

    The original function had singularites at z = -2 and z = 1, we subtracted from that a fiunction that also has singularities at z = -2 and z = 1, so the resulting function can only have singularites there, not at other points. However, since we have subtracted from the original function exactly the singular behavior of the function at the two points, the resulting function cannot have any singularities anymore. So, the conclusion is that

    z/[(z-1)(z+2)] - 2/3 1/(z+2) - 1/3 1/(z-1)

    is a rational function without any singularities. That implies that this function has to be a polynomial (degree of the denominator has to be zero). Since the fiunction clearly tends to zero at infinity, this means that the function is in fact zero everywhere. So, we have:

    z/[(z-1)(z+2)] - 2/3 1/(z+2) - 1/3 1/(z-1) = 0 ----->

    z/[(z-1)(z+2)] = 2/3 1/(z+2) + 1/3 1/(z-1)

    So, we have obtained the partial Fraction Expansion using Laurent expansions without having to solve any equations. The full Laurent expansion around z = -2 can now be obtained more easily by expanding 1/(z-1) around z = -2.

    Now I said that there are two Laurent expansions around z = -2. The other one is obtained by expanding 1/2 1/(z-1) around z = -2 in negative powers of z + 2. That expansion is also called "a Laurent expansion around z = -2", but, it will only converge for
    |z+2| > 3
     
  25. Apr 17, 2010 #24
    Okay,so I gather that the basic process is that we get the function,locate the poles....we separate out the singular part by partial fractions (in this case 2/3 1/(z+2))..and we put
    z/[(z-1)(z+2)] = 2/3 1/(z+2) + 1/3 1/(z-1) .....(1)

    ....leaving the singular part alone,we can now expand the 1/3 1/(z-1) as a normal Taylor series about -2 (because if we draw a circle around -2,we find that this part of the partial fraction does not have any singularity in the neighbourhood).....this expansion when put back into eqn (1) simply gives us the laurent series.

    So,if I'm okay till now,it implies that the laurent series isn't really anything special....we could extend this for higher powers of (z+2) in the denominator,and we would get more terms in the 'negative power part' of the laurent series...which appears to be nothing but a combination of partial fractions and taylor series.

    However,this doesn't explain the process they followed in my book where they substitute z+2 by t....and so the z+1 becomes (1-t)^(-1)....here,the singularity remains,as when z tennds to -2.t tends to zero.
     
  26. Apr 18, 2010 #25
    Okay,I've been thinking,and I read Count Iblis' post on https://www.physicsforums.com/showthread.php?t=49430 (#11) over again.....I guess that the fact that we can't get a Taylor series around a singularity simply means that we can't get a polynomial i.e positive powers of the variable around it....however,we can get an expansion around the singularity,which is not in positive powers....and this expansion is called the Laurent series,which we can obtain by manipulating the function we want to expand so as to get no singularities (paritial fractions,for example)...and now if we use the Taylor seires,and then reconvert the expansion in terns of the original variable,we get the negative power series ,

    If noone has any further comments on this,then, I presume my understanding is correct.
     
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