Integrating the Complex conjugate of z with respect to z

Deevise
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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

\int\bar{z}dz

where z is a complex number and \bar{z} is it's conjugate. Hmm i can't get the formatting to work out properly.. :S
 
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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)).
 
Well now i feel kind of stupid... its line intergration, not contour integration :P the question reads:

Evaluate the integral:

\int( \bar{z} +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...
 
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)).
 
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.
 
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...
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