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February 20th, 2018, 10:04 AM  #1 
Member Joined: Jan 2016 From: Blackpool Posts: 92 Thanks: 2  Cauchy's triangle theorem
Today in my lecture we proved that if a holomorphic function f is closed inside a triangle then the line integral is equal to 0. I was wondering why is this not the case for the set of real numbers, since they are a subset of the complex numbers. Thanks!

February 20th, 2018, 10:39 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 1,790 Thanks: 922 
Real functions other than constant functions will never be holomorphic as the CR equations will never be satisfied. $f(z) = u(z)+i v(z),~v(z)=0$ $u_x = v_y = 0$ $u_y = v_x = 0$ $u_x = u_y = 0 \Rightarrow f(z) = c$ You can convince yourself the line integral of a constant function over a closed contour equals 0. Last edited by romsek; February 20th, 2018 at 10:44 AM. 
February 22nd, 2018, 11:49 AM  #3  
Math Team Joined: Jan 2015 From: Alabama Posts: 2,966 Thanks: 807  Quote:
For example, suppose A= 0, B= 1, and C= 2. You are integrating form 0 to 1 to 2, back to A. The integral is $\displaystyle \int_0^0 f(x)dx= 0$.  
February 25th, 2018, 04:57 AM  #4 
Member Joined: Jan 2016 From: Blackpool Posts: 92 Thanks: 2 
does this mean that if I am asked to evaluate the integral where \[\int_{T}Re(z)dz\] where T is the triangle with vertices [0,1,1+i] then this integral will be equal 0? I don't think this is true as the real part of the complex function is not holomorphic which is a requirement for cauchys integral theorem? Thanks! 

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