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Integrating the Complex conjugate of z with respect to z

  • Thread starter Deevise
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  • #1
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Im doing a bit of contour integration, and a question came up with a term in it am unsure of how to do: in its simplest form it would be

[tex]\int[/tex][tex]\bar{z}[/tex]dz

where z is a complex number and [tex]\bar{z}[/tex] is it's conjugate. Hmm i can't get the formatting to work out properly.. :S
 
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Answers and Replies

  • #2
Dick
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If you are integrating over a circular contour of radius R then zz*=R^2, so z*=R^2/z. Otherwise you just have to take the contour and write it as z=(x(t)+iy(t)), so z*=(x(t)-iy(t)).
 
  • #3
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Well now i feel kind of stupid... its line intergration, not contour integration :P the question reads:

Evaluate the integral:

[tex]\int[/tex]( [tex]\bar{z}[/tex] +1 ) dz
L

Where L is the line segment from -i to 1+i.

normally i would just integrate and sub in start and end point, but i have totaly drawn a blank on what to do with the conjugate in this case...
 
  • #4
Dick
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Just treat it as a complex line integral. You can only 'sub in' endpoints if the function you are integrating is analytic and has an antiderivative. (z*+1) doesn't. Parametrize L as a function of t and integrate dt. Like I said, if you have z=(x(t)+iy(t)) then z*=(x(t)-iy(t)).
 
  • #5
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  • #6
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I think it's time i went to sleep... Yeh now that you mention the lack of anti-derivative i knew that. I think a good nights sleep will prepare me better for this exam than grinding my head into non-exsistant problems...

sorry to waste your time with inane questions lol... Thanks for the prompt responces.
 

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