Contour integrals

  • #1

Homework Statement



Calculate the following contour integrals [itex]\int_{c1} (x^3-3xy^2 ) + i (3yx^2 - y^3)[/itex] where c1 is th line from 0 to 1+i

Homework Equations





The Attempt at a Solution



a earlier part of the question asked if it was analytic. using Cauchy-Reimann equations i have shown it is. the next part asks me to calculate the contour integral.

so the problem is in the format u(x,y)+i v(x,y).

so is my integral = [itex]\int_{c1} u dx - \int_{c1} v dy + i[\int_{c1} u dy \int_{c1} + v dx[/itex]
if it is this, it seem straight forward enough, but im not sure then how to use the limit 0 to 1+i
am i going down the right road at all?
 

Answers and Replies

  • #2
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so is my integral = [itex]\int_{c1} u dx - \int_{c1} v dy + i[\int_{c1} u dy \int_{c1} + v dx[/itex]

You can parameterize c1 as (x,y)=(l*cos(pi/4), l*sin(pi/4)), so that dx=dl*cos(pi/4) and dy=dl*sin(pi/4), where l goes from 0 to sqrt(2). Alternatively, you can deform c1 to (0,0)->(0,1)->(1,1), so that dy=0 on the first arm and dx=0 on the second arm.
 
  • #3
Hi sunjin09,
Thanks a mill. its the parameterization that is throwing me. i dont understand where you get that from. is there anywhere you know i could look about it?
 
  • #4
fzero
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There's an easier parametrization. Since the contour is a line, x and y must be linear functions of a variable t. We can choose the endpoints as t=0 and t=1. Solve for the coefficients in the linear functions using the endpoints.
 
  • #5
312
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Hi sunjin09,
Thanks a mill. its the parameterization that is throwing me. i dont understand where you get that from. is there anywhere you know i could look about it?

A contour integral is nothing but a line integral on the complex plane. A parameterization of a line is given by z(t)=x(t)+iy(t), where t is a parameter, z(0)=start point and z(T)=end point. ( Actually this is the DEFINITION of a curve, as a continuous mapping from a closed interval of R to a closed set on the complex plane.)

The parameterization of the contour c1, which is a straight line from 0 to 1+i, is given by z=x+iy=r*(cos(pi/4)+i*sin(pi/4)), where r=|z| is the parameter. So dz=dx+idy=dr*(cos(pi/4)+i*sin(pi/4)), therefore dx=dr*cos(pi/4) and dy=dr*sin(pi/4). Sorry I just changed my notation from l to r.
 

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