Analyzing the Circulation Problem: Is There a Simpler Solution?

[Nikitin]

Homework Statement

https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-prn2/1395894_10201699046814591_1790916582_n.jpg

Assignment 3.

Homework Equations

Cauchy's differentiation formula
http://en.wikipedia.org/wiki/Cauchy's_integral_formula

The Attempt at a Solution

I checked if $f(z) = e^{-z}$ is analytic, and it is not. So it seems I have to do the circulation the hard way: parametrize z into $z(t) = \cos(t) +i \sin(t)$

But that sounds like a bad way to do it. So you guys got any better ideas? Maybe I am missing something?

thanks for all help! :)

 

[Office_Shredder]

Why is e^-z not analytic?

 

[Nikitin]

Oh right. Now that I have rested a bit at home I see it is obviously analytic. how stupid of me.

Anyway, how do I solve it?

I guess the strategy should be something like this?

1) Apply Cauchy's differentiation formula, with $z_0 = 0$ and $f(z_0) = e^{-z_0}$
2) Rearrange so the integral in assignment 3 is a function of $f^{(n)} (z_0)$.
3) Find $f^{(n)}$.

Err, how do I do step 3? Won't $f^{(n)}$ vary between being zero and 1 after inserting $z_0 = 0$, depending on the value of n?

 

[Dick]

Oh right. Now that I have rested a bit at home I see it is obviously analytic. how stupid of me.

Anyway, how do I solve it?

I guess the strategy should be something like this?

1) Apply Cauchy's differentiation formula, with $z_0 = 0$ and $f(z_0) = e^{-z_0}$
2) Rearrange so the integral in assignment 3 is a function of $f^{(n)} (z_0)$.
3) Find $f^{(n)}$.

Err, how do I do step 3? Won't $f^{(n)}$ vary between being zero and 1 after inserting $z_0 = 0$, depending on the value of n?

You are making this sound a lot more elaborate than it is. Just do it. Try a few examples, like n=1 and n=2 first. Sure $f^{(n)}(0)$ will depend on n. But it doesn't vary between zero and one.

 

[Nikitin]

Yes, you're right.

But I end up with

$$\frac{2 \pi}{(n-1)!} \cdot (-1)^{n-1} i$$

while the correct answer is just

$$\frac{2 \pi}{(n-1)!}$$

 

[Dick]

Yes, you're right.

But I end up with

$$\frac{2 \pi}{(n-1)!} \cdot (-1)^{n-1} i$$

while the correct answer is just

$$\frac{2 \pi}{(n-1)!}$$

That 'correct answer' doesn't look correct to me. I agree with your answer.

 

[Nikitin]

OK, thanks! :) I guess my book's solution-appendix is wrong on this assignment.