How Do You Evaluate Complex Integrals Using Taylor Series and Residues?

[Milky]

Homework Statement

Evaluate \oint_C \f(z) \, dz where C is the unit circle at the origin, and f(z) is given by the following:

A. e^{z}^{2}
(the z2 is suppose to be z squared)
B. 1/(z^{2}-4)

Homework Equations

The Attempt at a Solution

I'm completely confused

 

[malawi_glenn]

okay

You know this theorem:

if f(z) is analytic inside and on a closed loop C; then the integral of f(z) around the loop is zero

is e^{z^2} analytic inside and on the unit circle?

b. write z^2 - 4 as (z-2)(z+2); is 1/(z^{2}-4) analytic inside and on the unit circle?

 

[Milky]

Okay, so for

a. I get that e^{z^2} coincides with the unit circle at 1. But isn't it analytic at 1? Shouldn't the integral be zero? I'm not quite sure where to go from here.

b. I get that the integral is zero because the singularities are outside of the unit circle.

 

[malawi_glenn]

e^{z^2} is analytic on and inside the unit circle.

Infact e^{z^2} is entire, i.e analytic in the whole complex plane.

You should not calculate what the funtions are at certain values of z, you should just find out where they are analytic and where they are not. Then you use Cauchys integral theorem and other theorems (in the case when you have closed loops).

 

[Milky]

Okay, so I have a similar problem \oint_C \frac{e^{iz}}{z(z-\Pi)}\, dz

Here the singularities are z=0, and z= \Pi
Find the integral where
a. C is the boundary of the annulus between |z|=1 and |z|=4. So, this C encloses z=\Pi
b. C is the circle of radius R where R<\Pi. So, this C encloses z=0.

 

[malawi_glenn]

\pi

Gives you the "pi" in tex =)

And use \dfrac{nominator}{denominator}

To make nice fractions

Just so we can see exactly how the integral looks like.

use this:
http://en.wikipedia.org/wiki/Cauchy's_integral_formula

 

[Milky]

Okay I think that's about as close as I'm going to get lol

 

[malawi_glenn]

no just \pi

\Pi gives you the big pi =)

The trick is to for each case to identify the f(z) and the "a", then you just take the derivative of f(z) and evaluate it at z=a; then multiply with 2 \pi i in this case, since that factor is excluted in the problem, but must be there to use the formula.

Do you have a textbook on complex analysis?

 

[Milky]

Yes I do - complex variables by fokas and ablowitz, but I don't find it very helpful.

 

[malawi_glenn]

Yes I do - complex variables by fokas and ablowitz, but I don't find it very helpful.

okay, Saff and Snider is awesome I think.

But now, what do you think about the integral, can you give it a try and if you get stuck post exactly what you did and thought here so we can help you.

 

[Milky]

Okay.
The one I've tried so far is the one where C is the circle of radius R such that R> \pi. So, this circle would contain both singularities.

So,
\oint_C \frac{e^{iz}}{z(z-\Pi)}\, dz

= \oint_C \frac{\frac{e^{iz}}{z}}{(z-\pi)}\, dz + \oint_C \ \frac{\frac{e^{iz}}{z-\pi}}{z}\, dz

= 2i\pi \ (\frac{e^{iz}}{z}) + 2i\pi \ (\frac{e^{iz}}{z-\pi}) = 2i\pi \ (\frac{e^{iz}}{\pi} + \frac{e^{iz}}{-\pi}) = -4i

 

[Milky]

For the one where C is the boundary of the annulus between |z|=1 and |z|=4, and therefore encloses \pi, I get:\oint_C \frac{e^{iz}}{z(z-\Pi)}\, dz
=
\oint_C \frac{\frac{e^{iz}}{z}}{(z-\pi)}\, dz
= 2i\pi (\frac{e^{iz}}{z} = 2ie^{i \pi} = -2i

(Ugh, I feel like I'm doing something very wrong)

 

[Milky]

Finally, for the one where C is the circle of radius R such that R<\pi, I get:


\oint_C \frac{e^{iz}}{z(z-\Pi)}\, dz

= \oint_C \frac{\frac{e^{iz}}{z-\pi}}{(z)}\, dz

= 2i\pi (\frac{e^{iz}}{z-\pi} = 2i\pi \frac{e^{i\pi}}{-\pi} <br /> <br /> = 2i

I am not sure if I am the only who cannot see the final solution to this problem, but in the end i get 2i

 

[narjishanum]

i think the one for the circle of radius greater than pi is right...

 

[malawi_glenn]

According to the problem you told me in post #5 you had only two loops. One that encloses \pi and one that encloses 0.

a.
\oint_{C1} \frac{e^{iz}}{z(z-\pi)}\, dz

let C1 be curve that encloses \pi but not 0.

\oint_{C1}\frac{\frac{e^{iz}}{z}}{(z-\pi)}\, dz = 2 \pi i \frac{e^{i \pi}}{\pi} = -2i

b.
\oint_{C2} \frac{\frac{e^{iz}}{z-\pi}}{(z)}\, dz


let C2 be curve that encloses 0 but not \pi.

\oint_{C2} \frac{\frac{e^{iz}}{z-\pi}}{(z)}\, dz = 2 \pi i \dfrac{e^{0}}{0-\pi} = -2i

If you had a big loop that encloses both singularites; then you use the decompisition theorem of loops, and simply add upp the result from the two separate integrals =)

i.e -2i + (-2i) = -4i

Also the direction of the curve matters, if you go positive or negative.

 

[Milky]

I almost had it! Thank you so much for your help.
Also, does that mean that post number 11 is right?

 

[futurebird]

use this:
http://en.wikipedia.org/wiki/Cauchy's_integral_formula[/QUOTE]

I don't know if we can use that. It's in the next section. The problem seems to want us to use taylor series. I tried this but it made a big mess.

Is there a way to do the problem without Cauchy's integral formula?

 

[Milky]

Really? I used it for all of them - does that mean that my homework is completely wrong?

 

[Milky]

I have no idea how to do it using the Taylor series

 

[futurebird]

I have no idea how to do it using the Taylor series

I don't know, to be honest. Initially I tried substitution, but I ended up with something impossible to integrate using normal techniques. You can integrate some difficult expressions using taylor series and he mentioned that in class, and it's mentioned in the problem where it references the part of the book with all of the taylor series.

The thing is I can't find an example anywhere of a problem that uses taylor series for something as complex as this. I'm also fuzzy on how or where the series needs to be "centered" --

As the hour is growing late I worked out my answers using the formula, but I'm still trying to find a way to do it without the formula.

 

[Milky]

Well, hopefully he will accept what we did - and maybe he'll be proud that we read ahead (or maybe not lol). But I don't know how to do it the other way and I do remember him mentioning using either the taylor series and the power series - integrating term by term but - since I don't know how to do it I had no choice but to do it this way

 

[malawi_glenn]

futurebird; this is complex analysis. And they way to solve these integrals is by using Cauchys integral theorem. This is just basic integration in complex analysis.


milky: You did not take use of cachys integral formula in your post #11; you got the right result but not in the correct way.

I think that Cauchys integral formula is one of the coolest results in complex analysis, and also the theory of residue's that is a combination of cauchy integral formula and Laurent series.

If you wanted to solve these integrals by Taylor expansions and so on, then it should be adressed in the first post in the template "Know formulas" or whatever it is called again, so you could get proper help.

 

[futurebird]

If you wanted to solve these integrals by Taylor expansions and so on, then it should be adressed in the first post in the template "Know formulas" or whatever it is called again, so you could get proper help.

Can we address this anyway? I'm very curious.

I assumed #11 was using the integral formula. Otherwise how would you do that?

Let me type up what I've done.

 

[malawi_glenn]

When you start a new thread in one of the "Homework" forums; this templete will come:



So therefore, try using that 2. Writing what relations you know of and so on (not all, but some)

Now Futurebird, are you and Milky buddies?

And yes, post #11 he tried using Cauchy integral formula but did some errors.

Okay.
The one I've tried so far is the one where C is the circle of radius R such that R> \pi. So, this circle would contain both singularities.

So,
\oint_C \frac{e^{iz}}{z(z-\Pi)}\, dz

= \oint_C \frac{\frac{e^{iz}}{z}}{(z-\pi)}\, dz + \oint_C \ \frac{\frac{e^{iz}}{z-\pi}}{z}\, dz

= 2i\pi \ (\frac{e^{iz}}{z}) + 2i\pi \ (\frac{e^{iz}}{z-\pi}) = 2i\pi \ (\frac{e^{iz}}{\pi} + \frac{e^{iz}}{-\pi}) = -4i

here you should have:

C is a loop containing both pi and zero. and is positive oriented.
Ca is the loop containg just pi and Cb contain only 0; this we can do by the decomposition of loop/curves theorem. Ca and Cb are positive oriented.

\oint_C \frac{e^{iz}}{z(z-\Pi)}\, dz =

\oint_{Ca} \frac{\frac{e^{iz}}{z}}{(z-\pi)}\, dz + \oint_{Cb} \ \frac{\frac{e^{iz}}{z-\pi}}{z}\, dz =<br /> Now identify the different f(z)'s and evaluate them at z = 0 and z= pi, then multiply the whole thing with 2 pi i :

2i\pi \ (\frac{e^{i \pi}}{\pi}) + 2i\pi \ (\frac{e^{i*0}}{0- \pi}) = -4i

 

[futurebird]

\oint_{C}\frac{e^{iz}}{z(z-\pi)}dz

C is the circle around z =0

Let
z=re^{i\theta} and
dz=ire^{i\theta}d\theta

subbing that in:

\int^{2\pi}_{0}\frac{e^{rie^{i\theta}}ire^{i\theta}}{re^{i\theta}(re^{i\theta}-\pi)}d\theta

\int^{2\pi}_{0}\frac{ie^{rie^{i\theta}}}{re^{i\theta}-\pi}d\theta

Now I tried using this:

e^{z}=\displaystyle\sum_{j=0}^{\infty}\frac{z^{j}}{j!}

to replace:

e^{rie^{i\theta}}

But it looks like too much of a mess to work with term by term.

\int^{2\pi}_{0}\frac{1+ire^{i\theta}+\frac{-r^{2}e^{2i\theta}}{2!}+\frac{-ir^{3}e^{3i\theta}}{3!}+...}{(re^{i\theta}-\pi)}d\theta

 

[Milky]

I apologize, the reason I didn't enter that into the template is because it didn't occur to me that we needed to do it using the Taylor Series until Futurebird said it - my mistake. But that is completely fine with me because now I know how to use the Cauchy Integral Formula when I'm almost sure I wouldn't have understood how to use it otherwise. Thank you for your help.

I am also very curious and would like to know how to solve it using the Taylor series. I will try it later and post my solution. Any help would be appreciated :). Thanks again.

 

[malawi_glenn]

I don't remember how to use the Taylor series to do these integrals to be honest. Scince the cauchy formula is both simpler and more elegant, always use that one.

But anyway:
taylor.

If f is analytic at t, then:
f(t) + f&#039;(t) (z-t) + \frac{f&#039;&#039;(t)}{2!}(z-t)^2 + \frac{f&#039;&#039;&#039;(t)}{3!}(z-t)^3 = . . . = \sum_{j=0}^{\infty} \frac{f^{(j)(t)}}{j!}(z-t)^j

is called the taylor series for f around t.
So you first expand around t = 0 and then around t = pi, but as I said, I don't remember this method so i might get it wrong.

First to partial fractions of 1/(z(z-pi)); then you just stick in taylor for e^iz

(are you supposed to know decomposition of loops??)
(and may you use this:
\oint_{\gamma}\frac{1}{z-a}dz = 0
If gamma is closed loop and a is outside gamma, and:
\oint_{\gamma}\frac{1}{z-a}dz = 2 \pi i if gamma is closed loop and a is inside gamma. ?)

Then you get:

e^{iz}[ A/z + B/(z- \pi) ] = A[e^{i0} +ie^{i0}(z-0) + ... ]/z + B[e^{i \pi} +ie^{i \pi}(z-\pi) + ... ]/(z-\pi) =

\oint_{Ca}\frac{A}{z}dz + \oint_{Cb}\frac{B}{z - \pi}dz + \oint_{C&#039;} G(z) dz
Where Ca is a loop that contains 0, Cb a loop that contains pi; all positve oriented.
G(z) is a function that is entire, C' is a loop that contain both 0 and pi , so its integral becomes zero.

Now fil in the details, and do the proper partial fractions, then you'll probably have it