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February 20th, 2018, 10:04 AM  #1 
Member Joined: Jan 2016 From: Blackpool Posts: 97 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: 2,040 Thanks: 1063 
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: 3,261 Thanks: 894  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: 97 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! 
March 24th, 2018, 03:31 AM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
No. You originally said "I was wondering why is this not the case for the set of real numbers, since they are a subset of the complex numbers." The set of real numbers, as a subset of the complex numbers is the real axis. The "triangle theorem" in the real numbers is just, as I said, integration from point "A" on the real axis, out to point "B" on that axis, then back to "A". Such an integral is 0. The example you give is NOT "in the real numbers" because you are allowing z to take on nonreal values. And, of course, "Re(x)" is NOT a holomorphic function. 

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