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Question about Complex Integrals

  1. Nov 16, 2004 #1
    Hey there, I've just a quick question about numer 6 at the following link -

    http://titus.phy.qub.ac.uk/group/Jorge/AMA203/assignments/ass03_7.pdf [Broken]

    I'm normally okay with integrals but I'm a bit rusty on them at the mo (I haven't done a proper integral since May/June time, ooops!) and I'm convinced I'm doing something wrong here. For the first part (where the semicircle is in the upper half of the imaginary plane), I've let z=e^(itheta) so z*= e^(-itheta). Then to change the variable I've done dz/dtheta and got ie^(itheta). Since it moves around the upper half from x=1 to x= -1, I thought this would be equivalent to it going from 0 to pi. So I've calculated the integral (bottom limit=0 and upper limit=pi) and got i(pi) out of it.

    For the next one I wasn't sure if I was doing it right because it was going below the plane. I kept z=e^(itheta) since although the path was changing, I figured that since the angles we'd now be using would be negative, we could keep what looks like the same z as before. So I calculated it in the same way as the first part, except having my lower limit as 0 and my upper limit as -pi. In the end I got -i(pi) out of it.

    Can anyone check if those are okay or not? I guess I got confused at first because I expected thm to be equal, but then I think that z* isn't analytic so this isn't necessarily the case. If anyone could shed any light on it I'd be really grateful. Thanks. :smile:
    Last edited by a moderator: May 1, 2017
  2. jcsd
  3. Nov 17, 2004 #2
    It does make sence, the fact that z* is not analytical => it depends on the path.
  4. Nov 17, 2004 #3


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    But that doesn't mean two different paths can't give the same integral!

    Since both of these are on parts of the unit circle, convert to polar form:
    z= re so on the unit circle, z= e= cos(θ)+ i sin(θ) and z* (am I correct that that is the complex conjugate of z?) = cos(θl)- i sin(θ).

    The integral around the upper semi-circle is the integral of that from θ= 0 to θ= π and the integral around the lower semi-circle is the integral of that from θ= 0 to θ= -π.
  5. Nov 17, 2004 #4
    I never said that :P. Of course they can, but if is not analytical, there must be some (open) path where the integral is not the same, right?
  6. Nov 17, 2004 #5
    Yeah, I think that's the way I did it. I hope it's right anyway cos I handed it in before lunch today. :eek:
  7. Nov 17, 2004 #6
    Thanks, btw. :smile:
  8. Nov 17, 2004 #7
    Btw, while I'm in the mood for asking dumbo questions, we got our new h/work today and I'm a bit fuzzled by something. We're using he Darboux inequality to find out the maximum value of |f(z)| on C where C : |z| = 3 (the circle with radius 3). We need to know the length of C for this which I get to be 6pi just from using that the circumference of the circle is 2pir and r here is 3. However, if I wanted to work it out using an integral, I keep getting my answer to be zero!

    I start by putting z=3e^(itheta) and then having that dz= 3ie^(itheta)dtheta. I know that the length of C is the integral over dz, so I tried integrating what I got for dz between 0 (lower limit) and 2pi (upper limit), and all I can get is 0. I'm totally baffled. I mean I think I can see what the obvious answer is, but I just want to show it with integration........
  9. Nov 17, 2004 #8
    can you remind me of Darboux inequality pls?
  10. Nov 18, 2004 #9
    Sure, although I don't know how to use the proper maths typing that you lot use (technophobe!).

    It's given by | closed line integral of f(z) | <= ML . Or given in words (which might make a little more sense judging by my awful attempt to write that out properly), it's the modulus of the line integral of f(z) around the curve C, which is less than or equal to ML where M is the maximum value of |f(z)| on C and L is the length of C. So from the h/work I need to work out the values of M and L. At least I think that's what we've got to do. Here's the homework sheet it's taken from -

    http://titus.phy.qub.ac.uk/group/Jorge/AMA203/assignments/ass03_8.pdf [Broken]

    and it's queston 2.

    I'm also presuming that to work out M which is the maximum value of the modulus of f(z) on C we've to work out df/dz and set it equal to zero. Or something like that! I'm hopign I can do that okay, but I'm just confused about working out M using integrals, since I'm getting zero here where I should be getting 6pi.
    Last edited by a moderator: May 1, 2017
  11. Nov 18, 2004 #10
    So? whats your problem? You know the lenght of the curve right? The only thing you need to do is find the maximum of the norm over the curve.
  12. Nov 19, 2004 #11
    Yeah, I know that I know the length of the curve, but I want to show this length by integration and I don't know why it won't work. I'm guessing it's because the function I'm integrating is analytic but I'm not sure.

    The thing is, when you're finding the maximum for saya function such as f(z)= 1/((1+z^2)^3) then how can you do the differential to find the maximum? All you'll get is z=0 will you not (which doesn't lie on C)? I mean I've looked further on in my notes that we can write the modulus of the maximum of this f(z) as being less than or equal to the modulus of 1/((1-R^2)^3) since R is so much larger than 1 and z=Re^itheta...... I'm just not sure where all that comes from as in how you can change the modulus to be like that, like how can you be sure that's the maximum?....). I'm okay if we can set the differential equal to zero but what about if it results in this?
  13. Nov 19, 2004 #12


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    Integrating dz over a contour does not give the length of the contour. You want to integrate |dz| over the contour (notation varies, |dz|=ds, where s is arclength). If [tex]\gamma(t)[/tex] from [a,b] is a parameterization of your contour, then

    [tex]\int_{\gamma}|dz|=\int_{a}^{b}|\gamma '(t)|dt[/tex]

    You can try this on your circle, but it won't be a very exciting computation.

    To maximize |f(z)| on your circle, we can look to minimize [tex]|1+9e^{2i\theta}|[/tex] (after setting [tex]z=e^{i\theta}[/tex]). This is the same as minimizing [tex]|1+9e^{2i\theta}|^2=(1+9e^{2i\theta})(1+9e^{2i\theta})^{*}=82+18\cos{2\theta}[/tex], where * denotes the conjugate. Work out [tex]\theta[/tex] and you should see where the answer in your notes comes from. There's two choices of [tex]\theta\in[0,2\pi)[/tex] that give the max of |f(z)|.
    Last edited: Nov 19, 2004
  14. Nov 19, 2004 #13
    Thanks very much for your help there, it really helped esp with the maximizing. :smile: I've just realized I've been messing the others up, like if we had the integral of e^(2z) divided by z^2 + 1. Do we do the same, like by taking the modulus of it as above and then working out the min value? I can't believe it slipped my mind how to find the modulus of a complex number. It actually took me to look at my quantum mechanics notes to figure that one out. I'm so dozey! I think I'm going to head off to bed now cos I've been awake for 19 hours and I'm generally a pretty snoozy person! Thanks again.
  15. Nov 19, 2004 #14


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    You can work out the max of [tex]|e^{2z}/(z^2+1)|[/tex] the same way as before (dealing with the square of the modulus will make things simpler again). This will be a large pain though.

    If I wanted to find a bound for this integral, I would probably find the max of [tex]|e^{2z}|[/tex] on the circle and the max of [tex]|1/(z^2+1)|[/tex] on the circle, then multiply the values to get a bound for the modulus of the entire function. Using this bound rather than the maximum of the function won't give as tight a bound for the integral in the end, but not much will actually be lost and this kind of procedure is sufficient for most applications and *much* easier to apply.

    You might want to work out both methods to see the difference. You might also want to work out the exact value of the integral to see just how off these bounds actually are. The results might be disconcerting, but rest assured that these kinds of lousy looking bounds are often more than enough.
  16. Nov 20, 2004 #15
    I was just trying the first method and I wanted to know how we can find the conjugate of e^z? I know it sounds dozey but I really don't know. I mean I can do the conjugate of an ordinary complex number but not one where it's to the power of it. What we've got in our notes it stuff like (z1/z2)* gives us the conjugate of z1 over the conjugate of z2, but I don't kno how it works when we''ve got the exponential. Should the exponential be changed into another form?

    If I did it the second way I'd have said that e^z would have been maximum at theta = 0, 2pi etc (only guessing though, I'm not sure how you could work this one out exactly) , but for the first one I wasn't sure.....

    Btw, for the second bit when you say multiply the 2 bounds together, do you mean divide? Like since z^2 + 1 is on the denominator?

    Btw, sorry about all the questions, I guess I'm a bit dozey. Thanks again for your help - you're proving to be a lot more helpful than the recommended texts!
    Last edited: Nov 20, 2004
  17. Nov 20, 2004 #16
    Suppose z = x + yi, then conjugate(e^z) = conj(e^(x + yi)) = e^x * conj(e^(yi)) = e^x * conj(cos(y) + isin(y)) = e^x(cos(y) - isin(y)) = e^x(cos(-y) + isin(-y)) = e^x * e^(-yi) = e^(x - yi) = e^(conj(z)).
  18. Nov 20, 2004 #17


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    Your intuition is correct here. Use [tex]e^{a+ib}=e^{a}(\cos(b)+i\sin(b))[/tex] to work out the modulus of e^z on the circle, where [tex]a+ib=z[/tex].

    I meant maximum- sorry I did a change up and was talking about an upper bound for [tex]|1/(z^2+1)|[/tex]. If you had a lower bound for [tex]|z^2+1|[/tex], then you'd divide as you say. I should mention that you can use the triangle inequality to find a bound if you aren't concerned with the tightest bound possible. For example [tex]|z^2+1|\geq |z^2|-|1|=|z|^2-1=R^2-1[/tex] on the circle of radius R. This particular lower bound is only meaningful if R>1 though. In this case it does give the tightest bound, but this won't always happen.
  19. Nov 20, 2004 #18
    Ah right, so I can have that if we have e^z and I change that so z=re^(itheta), and (e^z)* will give us e^(e^-(itheta))?

    Btw, thanks for the heads up on the triangle inequality. :smile:
  20. Nov 20, 2004 #19
    Anyhoo, I've been faffing about with finding the tightest possible upper bound for the closed integral of the modulus of (e^2z)/(z^2 +1)

    I did the square of the modulus so I'm now looking at finding the maximum of (e^(12cos(theta)))/(82 + 18cos(2theta)) which I got after a bit of work, but now I'm a bit stuck and I'm not sure how much further to take it. I'm maybe seeing now why it'd be better to not have as tight an upper bound, but I'd just like to see if I can get this one to work. :smile:
  21. Nov 24, 2004 #20


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    Hi, you should be able to maximize that via the usual methods, differentiate w.r.t. theta, set=0, etc. It's a pain, but using some trig identities (a half angle formula and sin^2+cos^2=1 should do it), it should be feasible.
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