Complex integration Not quite "getting" it.
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Complex integration Not quite "getting" it.
I already understand how and why the CauchyRiemann equations work for Complex Differentiation (limit is valid for any direction in the plane, and the equations are the result of equating the derivatives taken from the real and imaginary axes), but as for its life partner, Complex Integration, I don't really get it. Due to not being in a Math course that actually covers any advanced stuff like this, I'm using the internet for information. Unfortunately, it's difficult to understand, most likely because I don't have background knowledge in Vector Calculus. Could anyone give me a bit better, perhaps a somewhat graphical way to explain complex integration?
 scarecrovv
 It's pronounced 'double u'
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Re: Complex integration Not quite "getting" it.
I am not a mathematician. I have never done complex integration. I top out at differential equations. If you know better than what I say, then ignore me. However, I would just define complex integration like this:
You have a continuous function f(t) that returns complex numbers given real numbers. It's like a path integral. If you trace out a path in the complex plane (a+bi) where da/dt and db/dt determine the direction and magnitude of the step for every dt, the answer is the complex number you need to add to the start point to get to the end point.
You have a continuous function f(a+bi) that returns real numbers given complex numbers. It's like a double integral, and it returns the volume of the solid bounded by the complex plane, the surface the function defines over it, and the bounds of integration.
You have a continuous function f(a+bi) that returns complex numbers given other complex numbers. Some funny fourdimensional thing happens???? Profit!
Remember, I have no idea what I'm talking about, and if you know what the CauchyRiemann equations are, you know more than I do about this.
You have a continuous function f(t) that returns complex numbers given real numbers. It's like a path integral. If you trace out a path in the complex plane (a+bi) where da/dt and db/dt determine the direction and magnitude of the step for every dt, the answer is the complex number you need to add to the start point to get to the end point.
You have a continuous function f(a+bi) that returns real numbers given complex numbers. It's like a double integral, and it returns the volume of the solid bounded by the complex plane, the surface the function defines over it, and the bounds of integration.
You have a continuous function f(a+bi) that returns complex numbers given other complex numbers. Some funny fourdimensional thing happens???? Profit!
Remember, I have no idea what I'm talking about, and if you know what the CauchyRiemann equations are, you know more than I do about this.
Re: Complex integration Not quite "getting" it.
Ignore what scarecrovv said, no offense intended. It's kind of along the right lines, but misses the nature of what's going on.
Think about what a complex function f(z) does. The complex plane is, in a sense, twodimensional, and any function w = f(z) that maps from C to C (too lazy to look up the LaTeX) is somewhat of a different beast than a real valued function g(x,y). g(x,y) gives you a surface. f(z) does not. Real valued functions can be thought of as giving a "height": h(x) gives the height of the curve at every x; g(x,y) gives the height of the surface at every (x,y). But f(z) can give complex values. How high is 3+7i? It doesn't mean anything. Complex functions are mappings. If you're familiar with the Riemann sphere, think about f(z) as a set of rules of how to twist and stretch a sheet that covers the Riemann sphere. Some z gets mapped to some new complex value w.
So what does it mean to do complex integration? We don't do complex integration over some domain, since that doesn't really make a whole lot of sense. Instead, we do it along a curve. If this curve is closed, then what we have is analogous to a line integral in realvalued calculus.
If you write dz = dz + idy, and invoke Green's Theorem and CauchyRiemann, then you'll see that any contour integral of a complex function that is analytic everywhere inside your contour evaluates to zero. This PDF document explains it better than I can: http://www.maths.tcd.ie/~richardt/3E1/2 ... cauchy.pdf
What gets interesting is when the function has poles, i.e., points that go to infinity within the contour. Here, we cannot invoke Green's Theorem. Instead, we need to find the limiting case of the function at the poles as the contour closes around them. The reasoning behind this is fairly simple: the value of the integral can be shown to be dependent only on the singularities it encloses, and not on the actual contour itself. So we can collapse the contour around the poles, and take a limiting case, and find that the influence of all the stuff except for the poles is zero. What's left is a "residue" on the poles. This is what we can compute using the Residue Theorem.
Often times, the Cauchy Integral Formula is taught first. I think this is silly, because CIF is just a special case of the Residue Theorem, and can be easily obtained from it.
Think about what a complex function f(z) does. The complex plane is, in a sense, twodimensional, and any function w = f(z) that maps from C to C (too lazy to look up the LaTeX) is somewhat of a different beast than a real valued function g(x,y). g(x,y) gives you a surface. f(z) does not. Real valued functions can be thought of as giving a "height": h(x) gives the height of the curve at every x; g(x,y) gives the height of the surface at every (x,y). But f(z) can give complex values. How high is 3+7i? It doesn't mean anything. Complex functions are mappings. If you're familiar with the Riemann sphere, think about f(z) as a set of rules of how to twist and stretch a sheet that covers the Riemann sphere. Some z gets mapped to some new complex value w.
So what does it mean to do complex integration? We don't do complex integration over some domain, since that doesn't really make a whole lot of sense. Instead, we do it along a curve. If this curve is closed, then what we have is analogous to a line integral in realvalued calculus.
If you write dz = dz + idy, and invoke Green's Theorem and CauchyRiemann, then you'll see that any contour integral of a complex function that is analytic everywhere inside your contour evaluates to zero. This PDF document explains it better than I can: http://www.maths.tcd.ie/~richardt/3E1/2 ... cauchy.pdf
What gets interesting is when the function has poles, i.e., points that go to infinity within the contour. Here, we cannot invoke Green's Theorem. Instead, we need to find the limiting case of the function at the poles as the contour closes around them. The reasoning behind this is fairly simple: the value of the integral can be shown to be dependent only on the singularities it encloses, and not on the actual contour itself. So we can collapse the contour around the poles, and take a limiting case, and find that the influence of all the stuff except for the poles is zero. What's left is a "residue" on the poles. This is what we can compute using the Residue Theorem.
Often times, the Cauchy Integral Formula is taught first. I think this is silly, because CIF is just a special case of the Residue Theorem, and can be easily obtained from it.

 Posts: 2
 Joined: Mon Feb 21, 2011 12:59 am UTC
Re: Complex integration Not quite "getting" it.
Big thanks to gorcee! The compare and contrast between integration in the complex plane vs the real function g(x,y) cleared up a major misconception.
Now an addition question: The example given at http://www.mathnotes.org/?pid=109#?pid=109 integral, int((exp(1+i)*x)dx = ((1i)/2)*(exp((1+i)*x)*(C1+iC2) appears to be simple(?) integration with due respect to treating i as a constant.
This example is worked out in full and I don't see the concept of a path in the complex plane entering into this particular example. I followed the example in Mathcad and, using the Mathcad symbolics, got to the same results (in painful detail).
What am I missing?
(Please forgive any missing or redundant parenthesis in the above. I am dyslexic when it comes to counting parenthesis.)
Thank you.
Now an addition question: The example given at http://www.mathnotes.org/?pid=109#?pid=109 integral, int((exp(1+i)*x)dx = ((1i)/2)*(exp((1+i)*x)*(C1+iC2) appears to be simple(?) integration with due respect to treating i as a constant.
This example is worked out in full and I don't see the concept of a path in the complex plane entering into this particular example. I followed the example in Mathcad and, using the Mathcad symbolics, got to the same results (in painful detail).
What am I missing?
(Please forgive any missing or redundant parenthesis in the above. I am dyslexic when it comes to counting parenthesis.)
Thank you.
Re: Complex integration Not quite "getting" it.
CasualObserver wrote:Big thanks to gorcee! The compare and contrast between integration in the complex plane vs the real function g(x,y) cleared up a major misconception.
Now an addition question: The example given at http://www.mathnotes.org/?pid=109#?pid=109 integral, int((exp(1+i)*x)dx = ((1i)/2)*(exp((1+i)*x)*(C1+iC2) appears to be simple(?) integration with due respect to treating i as a constant.
This example is worked out in full and I don't see the concept of a path in the complex plane entering into this particular example. I followed the example in Mathcad and, using the Mathcad symbolics, got to the same results (in painful detail).
What am I missing?
(Please forgive any missing or redundant parenthesis in the above. I am dyslexic when it comes to counting parenthesis.)
Thank you.
The difference here is that they're not doing a complex integral, they're doing a real integral.
In other words, even though the have a complex number floating around in the integrand, the variable of integration, x, is still a real number.

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 Joined: Mon Feb 21, 2011 12:59 am UTC
Re: Complex integration Not quite "getting" it.
Thanks for your prompt response. Opps; (1+i)*x is (x+i*x) not (x+ix). Burned by the parenthesis again.
Re: Complex integration Not quite "getting" it.
CasualObserver wrote:Thanks for your prompt response. Opps; (1+i)*x is (x+i*x) not (x+ix). Burned by the parenthesis again.
(x+i*x)
(x+ix)
These two things are the same =P
Re: Complex integration Not quite "getting" it.
Even though x+ix is complex, the "small increment" in the integral is dx, which is real. So our integral is not "really" a complex integral. If you have an integral that's a sum of "infinitely many" things of the form f(z)dz where f(z) and dz are both complex, then you're "really" doing a complex integral.
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