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 February 20th, 2018, 10:04 AM #1 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 104 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: 2,529 Thanks: 1389 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 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 104 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: 3,264 Thanks: 902 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 non-real values. And, of course, "Re(x)" is NOT a holomorphic function. Tags cauchy, theorem, triangle Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Howard Johnson Calculus 2 May 21st, 2016 06:31 PM WWRtelescoping Complex Analysis 2 April 22nd, 2014 02:51 AM bigli Complex Analysis 5 February 1st, 2012 05:51 AM stone Number Theory 0 July 13th, 2010 01:12 PM Seng Peter Thao Applied Math 0 June 30th, 2007 10:44 AM

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