Complex Integration: Contour Evaluation and Estimation Lemma

nickolas2730
Messages
28
Reaction score
0
1.Evaluate ∫C Im(z − i)dz, where C is the contour consisting of the circular arc along |z| = 1 from z = 1 to z = i and the line segment from z = i to z = −1.

2. Suppose that C is the circle |z| = 4 traversed once. Show that
§C (ez/(z+1)) dz ≤ 8∏e4/3

For question 1, should i let z= x+yi to solve the question?
and it said the I am part so i just need to consider the "yi"?

i tried but really have no idea on these 2 questions..
Please help
Thanks
 
Physics news on Phys.org
Yes, let z=x+iy and then integrate iy-i over z=e^(it) as t goes from 0 to pi/2. So wouldn't that be:

\int_0^{\pi/2} Im(z-i)dz,\quad z=e^{it}

You can convert that to all in t right? dz=ie^(it)dt=i(cos(t)+isin(t))dt and won't y be just sin(t)?

For the contour from i to -1, need to do that one in terms of z=x(t)+iy(t). Isn't that line just y=x+1 as x goes from 0 to -1? So suppose I let x=x(t)=t, then y(t) is? Now substitute all that into the integral:

\int_0^{-1} (iy-i)dz,\quad z=x(t)+iy(t)

with dz=x'(t)+iy'(t)
 
For the second one, use the http://en.wikipedia.org/wiki/Estimation_lemma" .
 
Last edited by a moderator:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top