# Complex Analysis Integral

1. Oct 10, 2007

### Milky

1. The problem statement, all variables and given/known data
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. $$1/(z^{2}-4)$$

2. Relevant equations

3. The attempt at a solution
I'm completely confused

Last edited: Oct 10, 2007
2. Oct 10, 2007

### 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?

3. Oct 11, 2007

### 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.

4. Oct 11, 2007

### 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).

5. Oct 11, 2007

### 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.

Last edited: Oct 11, 2007
6. Oct 11, 2007

### malawi_glenn

\pi

Gives you the "pi" in tex =)

And use \dfrac{nominator}{denominator}

To make nice fractions

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

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

Last edited: Oct 11, 2007
7. Oct 11, 2007

### Milky

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

Last edited: Oct 11, 2007
8. Oct 11, 2007

### 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?

9. Oct 11, 2007

### Milky

Yes I do - complex variables by fokas and ablowitz, but I dont find it very helpful.

10. Oct 11, 2007

### malawi_glenn

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 exaclty what you did and thought here so we can help you.

11. Oct 11, 2007

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

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

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

Last edited: Oct 11, 2007
12. Oct 11, 2007

### 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)

Last edited: Oct 11, 2007
13. Oct 11, 2007

### Milky

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

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

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

= $$2i\pi (\frac{e^{iz}}{z-\pi} = 2i\pi \frac{e^{i\pi}}{-\pi} = 2i$$

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

Last edited: Oct 11, 2007
14. Oct 11, 2007

### narjishanum

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

15. Oct 11, 2007

### 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.

Last edited: Oct 11, 2007
16. Oct 11, 2007

### Milky

Also, does that mean that post number 11 is right?

17. Oct 11, 2007

### futurebird

Last edited by a moderator: May 3, 2017
18. Oct 11, 2007

### Milky

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

Last edited: Oct 11, 2007
19. Oct 11, 2007

### Milky

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

20. Oct 11, 2007

### futurebird

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.