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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 C-R 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. Thanks from Jaket1 Last edited by romsek; February 20th, 2018 at 10:44 AM.
February 22nd, 2018, 11:49 AM   #3
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 Originally Posted by Jaket1 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!
Actually, it is true. Of course, since the real numbers lie on a line, and a triangle is a two dimensional figure, it becomes rather trivial. It is true in the sense that if you integrate from point A to point B, then point C, and back to point A, on a line you have integrated from A to either B or C, whichever is farther, then back to A. But the integral from A to A is always 0.

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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