# Homework Help: Q on countour integrals

1. May 20, 2007

### trickae

Contour Integral question

1. The problem statement, all variables and given/known data

http://img300.imageshack.us/img300/8536/20075202330586331530065ra7.jpg [Broken]
I need to evaluate the following integral over the above contour - also could someone do it for integral (ez cos z) as well?

2. Relevant equations

3. The attempt at a solution

I tried the method of deformation, i tried using the form

Integral [F(z) dz] = integral [f(z(t)) . z'(t)]

neither method works - because i get stuck near the end.

my working
:
Integral [F(z) dz] = integral [f(z(t)) . z'(t)]

the parametrized form of the contour - i used the deformation principle

that

http://img525.imageshack.us/img525/5272/myworkingcomplexq3wy8.jpg [Broken]

yet i doubt this is the method - any help?

Last edited by a moderator: May 2, 2017
2. May 21, 2007

### trickae

can someone explain this?

we know that ez sinz is an entire function. i.e. it has no singularities.and the region D is bounded by the simple closed path.
so by Cauchy integral theorem, we have http://img95.imageshack.us/img95/8707/cramsterequation2007521ed3.png [Broken]=[/URL] 0
i.e. http://img221.imageshack.us/img221/9658/cramsterequation2007521mb9.png [Broken]=[/URL] 0.
similarly http://img138.imageshack.us/img138/1315/cramsterequation2007521gx5.png [Broken] =0

Last edited by a moderator: May 2, 2017
3. May 21, 2007

### siddharth

4. May 21, 2007

### HallsofIvy

Cauchy's integral formula says that the integral of an entire function, over a closed contour, is 0. The contour in this problem is not closed.
It is, however, true that you can replace the given contour by any other contour having the same endpoints. You might use the quarter circle or two lines, from i down the imaginary axis to 0 and then up the real axis to $\pi$.

Last edited by a moderator: May 2, 2017
5. May 21, 2007

### trickae

if the contour isn't closed then in fact it isn't 0 then is it? or does the above still hold? I was thinking it was integrable ....

6. May 21, 2007

### HallsofIvy

If the contour is not closed then the integral is not necessarily 0 but might be. I'm not sure what you mean by "I was thinking it was integrable"- ezsin(z) obviously is integrable! As I said, you can use any contour having the same endpoints- in particular the quarter-circle you mentioned or two line segments. Yet another way to do this is to "ignore" the contour: find an anti-derivative of ezsin(z) and evaluate at i and $\pi$.

7. May 22, 2007

### trickae

thanks i'll do that.