1. Nov 16, 2004

Claire84

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.

Last edited by a moderator: May 1, 2017
2. Nov 17, 2004

ReyChiquito

It does make sence, the fact that z* is not analytical => it depends on the path.

3. Nov 17, 2004

HallsofIvy

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(&theta;l)- 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.

4. Nov 17, 2004

ReyChiquito

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?

5. Nov 17, 2004

Claire84

Yeah, I think that's the way I did it. I hope it's right anyway cos I handed it in before lunch today.

6. Nov 17, 2004

Claire84

Thanks, btw.

7. Nov 17, 2004

Claire84

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........

8. Nov 17, 2004

ReyChiquito

can you remind me of Darboux inequality pls?

9. Nov 18, 2004

Claire84

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
10. Nov 18, 2004

ReyChiquito

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.

11. Nov 19, 2004

Claire84

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?

12. Nov 19, 2004

shmoe

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 $$\gamma(t)$$ from [a,b] is a parameterization of your contour, then

$$\int_{\gamma}|dz|=\int_{a}^{b}|\gamma '(t)|dt$$

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 $$|1+9e^{2i\theta}|$$ (after setting $$z=e^{i\theta}$$). This is the same as minimizing $$|1+9e^{2i\theta}|^2=(1+9e^{2i\theta})(1+9e^{2i\theta})^{*}=82+18\cos{2\theta}$$, where * denotes the conjugate. Work out $$\theta$$ and you should see where the answer in your notes comes from. There's two choices of $$\theta\in[0,2\pi)$$ that give the max of |f(z)|.

Last edited: Nov 19, 2004
13. Nov 19, 2004

Claire84

Thanks very much for your help there, it really helped esp with the maximizing. 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.

14. Nov 19, 2004

shmoe

You can work out the max of $$|e^{2z}/(z^2+1)|$$ 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 $$|e^{2z}|$$ on the circle and the max of $$|1/(z^2+1)|$$ 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.

15. Nov 20, 2004

Claire84

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
16. Nov 20, 2004

Muzza

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)).

17. Nov 20, 2004

shmoe

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

I meant maximum- sorry I did a change up and was talking about an upper bound for $$|1/(z^2+1)|$$. If you had a lower bound for $$|z^2+1|$$, 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 $$|z^2+1|\geq |z^2|-|1|=|z|^2-1=R^2-1$$ 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.

18. Nov 20, 2004

Claire84

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.

19. Nov 20, 2004

Claire84

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.

20. Nov 24, 2004

shmoe

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.

21. Nov 26, 2004

Claire84

I handed it in with the more simple way so *fingers crossed* it comes out okay. I asked the lecturer the day before I was to hand it in to see if he could give me a bit of feedback on how tight the bound had to be, but he just told me that I was the mathmatician and what did I think...... not quite what I was hoping for but nevermind! Makes me thankful I found this place or sometimes I'm not sure how I'd get to grips with some of the material that we cover!

22. Nov 26, 2004

Claire84

Speaking of integrals, I was hopign someone could give me a hint as to what I've to do to get the integral between 0 and + infinity of (sin^2x)/(x^2). I can integrate the same thing just without the squares in it by doing the whole method of using a semicircle with a little arch in the middle to avoid the singularity at 0, but I can't do this here because there comes a point in it where I end up with a completely up the left answer (when we have to let r, the radius of the little arch, go to zero and we have the integral around the small circle of (e^i2z)/(z^2). Normally when you change the variable you're left with something v.easy to integrate, but not here as you still have an r on the bottom so it shoots to infinity). I read a formula online for this kinda thing where the closed integral of f(z) = 2pi i(sum of enclosed residues)+ pi i (residues on the boundary) which works fine for mme, but this isn't something we're being taught so I was hoping someone else could give me a push in the right direction about this. If you want to seemy working out for what I've done so far feel free to say, as I'll put it up on a word document since I can use (shock, horror!) equation editor on that.

23. Nov 26, 2004

Claire84

Decided just to write it up on word anyway so here's my working out (hope it appears!)

Attached Files:

• Consider the integral of the function.doc
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24. Nov 27, 2004

Claire84

Oh, and we've to show that it is equal to pi.

25. Nov 27, 2004

shmoe

Are you opposed to using the Laurent series of $$\frac{1-e^{2iz}}{z^2}$$ at z=0 to find it's integral on your path a? This is what the +pi*i*(residue on boundary) kind of results you found would have used. It's not too difficult once you've found the Laurent series and ties in very nicely with the work you've been doing on bounding certain integrals.

So I suggest find the series and try to integrate it. A hint: the 1/z term is to be trated seperately, the rest of the series can be dealt with in one fell swoop. More hints available on request.