Why Does the Complex Integral Not Represent Area Like in Real Calculus?

[blumfeld0]

Hi. I am studying on complex variables on my own using Brown and Churchill, 6th edition.
so far i am doing pretty good, but i have run into trouble doing a few contour integrals.

1. First off, the authors state that unlike in calculus, where the integral can mean area under the curve, in complex analysis, it does not have any physical meaning (except in rare special cases). this is what they claim. really? no physical meaning? i was very surprised to learn this. why isn't it just the area under the curve?

ok here are the problems i just can't get:

2. f(z) = Pi *exp(Pi z_) and C is the boundary of the square with vertices at the points 0,1, 1+i and i, the orientation of C is counterclockwise direction.
where z_ = x - i y.
i know I have to break up the integral into two. my limits of integration for both are going to be from 0 to 1 i believe.
the first one will be integral from 0 to 1 of (-i) * Pi *exp(Pi z_)
the second integral will be from 0 to 1 of (x+i) * Pi *exp(Pi z_)
then i add up the result. is that right?

3.

f(z) is the branch z^(-1+i) = exp((-1+i)*log[z]) ( |z|>0, 0<arg z <2 Pi)
and C is the positively oriented unit circe |z|=1.
I really thought i had this but my answer isn't right. (I have the solution)
I know log[z] = Ln r + i*theta
r = 1 so ln r = 0 so log[z] = i*theta

Integral from 0 to 2Pi of exp((-1+i)*i*theta) * (-1+i)*i
but it ain't workin'.

3. lastly,
f(z) is defined by the equations

f(z) = { 1 when y<0 and 4y when y>0}
and C is the arc from z = -1-i to 1+i along the curve y =x^3.

this i don't even know where to start.

thank you!

 

[mathwonk]

in complex analysis the integrals are all path integrals, and they absically are similar to the path integral of dtheta, the angle form, so their interpretation is something like winding number.i.e. integrals of differentiable functions give zero around a path, but integrands which are not defiend everywhere give an answer that depends on how many times your path winds around those places where the integrand is not defined.

(The graph of a complex function is a surface in 4 space so "area under the curve" makes no sense.)

 

[tacman]

Hello! It's great to hear that you are studying complex variables on your own. It can be a challenging but rewarding subject. I'll do my best to address your questions and help you with the problems you are struggling with.

1. Yes, it is true that in complex analysis, the integral does not have a physical meaning like it does in calculus. This is because the complex plane does not have a "direction" like the real number line does. In calculus, we can think of the integral as finding the area under a curve, but in complex analysis, the integral is more about the "net effect" of a function along a path. This can be hard to visualize, but with practice, you will get the hang of it.

2. For the first problem, your approach is correct. You have to break up the integral into two parts, and the limits of integration for both parts will be from 0 to 1. However, your integrands are not quite right. For the first integral, it should be (-i) * Pi * exp(Pi z_), and for the second integral, it should be (x+i) * Pi * exp(Pi z_). Remember that the variable z_ is equal to x - iy, so when you substitute it into the function, you have to be careful with the signs.

3. For the second problem, your approach is also correct. However, there seems to be a small error in your calculation. The integral should be from 0 to 2Pi of exp((-1+i)*i*theta) * (-1+i)*i. Remember to use the chain rule when you take the derivative of the logarithm. The correct answer should be (-1+i) * 2Pi.

4. Finally, for the third problem, you will need to use the Cauchy Integral Formula. This formula states that if a function is analytic (meaning it has a derivative) inside a closed contour, then the value of the integral around the contour is equal to the value of the function at a point inside the contour. In this case, you can use the point z = 0 as your point inside the contour. From there, you can break up the integral into two parts, one for the arc from -1-i to 0 and one for the arc from 0 to 1+i. You can then use the definition of the function f(z) to evaluate the integral.

I hope this