Question about complex integrals

  • Thread starter Thread starter LiamH
  • Start date Start date
  • Tags Tags
    Complex Integrals
LiamH
Messages
2
Reaction score
0

Homework Statement



Hey people got a question here about complex integration, not really sure how to do it so hope someone out there could help me!

Evaluate the complex integrals

∫ c { (zbar)^2 +1 } dz...and...∫ c { zcos(z^2) - ie^2z }

where c is the contour joining 0 to 2i along
(i) the line segment parametrized by z(t) = 2it, t∈{0,1}
(ii) the semicircle parametrized by z(t) = i + e^it, t∈{-pi/2, pi/2}



Homework Equations





The Attempt at a Solution


I know we have to sub the parameterization in for z, but now that there's a line and semicircle I am confused! Didnt have a great prof for this subject, so I am hopin some other people could help me out
 
Physics news on Phys.org
hi liamH

as you are given the parameterisaton, why not try substituting it into the inetrgal, then the integration should become a simple single variable integral. Remember you will have to account for the dz as well.
 
Ok so for the first integral and the line segment...

int from 0 to 1 {(2it bar) ^ 2 +1} (2i) dz

i worked this out and the answer i got was -2/3 i?

is this right?

and so for the semicircle the integral wil be

int from -pi/2 to pi/2 {(-1 + e^it bar) ^2 + 1 (e^it) } ?
 
first one looks correct - probably easier if you show you working and I can check it rather than me doing the whole integral as well
 
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