Couple questions about complex integrals

cAm
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we recently started integrals over the complex plane, in complex analysis. But, I am confused on the iterpretation of some of it. From what I've heard/understand, there Is no geometrical interpretation of a complex integral. Why is this? Also, why is it 'illegal' to do an indefinate integral. What's the difference between doing that, and an antiderivative?
 
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"From what I've heard/understand, there Is no geometrical interpretation of a complex integral. Why is this? Also, why is it 'illegal' to do an indefinate integral. "

I can't answer this because I've never heard either of those things. Certainly, a complex function of a complex variable would have to be represented in 4 dimensions. Perhaps that is what was meant- 4 dimension geometry is hard to imagine!
The indefinite integral of f(z)= 2z is z2+ C whether z is a real variable or complex. Nothing "illegal" about that!
 
HallsofIvy said:
"From what I've heard/understand, there Is no geometrical interpretation of a complex integral. Why is this? Also, why is it 'illegal' to do an indefinate integral. "

I can't answer this because I've never heard either of those things. Certainly, a complex function of a complex variable would have to be represented in 4 dimensions. Perhaps that is what was meant- 4 dimension geometry is hard to imagine!
The indefinite integral of f(z)= 2z is z2+ C whether z is a real variable or complex. Nothing "illegal" about that!


hmm... we've beent tought that indefinate integrals are undefined, though we can do antiderivatives.

And, do you know what the geometrical representation is, even if it is hard to conceptualize in 4d?
 
"hmm... we've beent tought that indefinate integrals are undefined, though we can do antiderivatives"

Indefinite integrals ARE anti-derivatives.
 
HallsofIvy said:
"hmm... we've beent tought that indefinate integrals are undefined, though we can do antiderivatives"

Indefinite integrals ARE anti-derivatives.


That's what i always thought, but my teacher is specifically saying that if we use an 'indefinate integral' then we'll lose some points. Whereas, an antiderivative is fine.
 
4d geometry is a good thinking exercise to try and visualise. Right, imagine a function that takes and returns a real number (ie it exists on the 'normal' number line). You can represent this as a line on a 2d graph, with aeis, say, x and y being at a 90^\circ angle to each other. In this 2 dimensional space, the axes x and y are said to be orthogonal - finding out the x position of a point gives you no information on the y position. You then extend this to three dimensions, we get the x, y and z axes. Again, knowledge of a point in 3D's position on one axis contains no information about the point's position on the other two axes - the three axes are again orthogonal.

Now think of this - you can project a 3D graph onto a 2D graph by taking a slice through at, say, a given Z value. For example, a sphere in 3d withh describe either nothing, a circle or a point when projected for a given 'slice' into 2D.

This all works nice and easy because we live in 3D. The conceptual leap comes when you consider 4 orthogonal axes. In a function taking and returning complex values we require 4 values to describe the function's behaviour:

f(z_{1}) = f(a+bi)= z_{2} = c+di

Where z_{1,2} are complex and a, b, c, d are real. These four numbers can represent four orthogonal axis, just as for f(x)=y, x and y represent values on two orthogonal axes.

Try to imagine slices through this 4D space. They will be able to be represented as 3D graphs that you can visualise.

I hope that helps - mull it over a bit.
 
Also, why is it 'illegal' to do an indefinate integral. What's the difference between doing that, and an antiderivative?

perhaps this question is in regard to the definition of the integral in the respect to Riemann sums. By the way, what would be the geometrical interpretation of a antiderivative F(a) on the graph of f(x)? I have an idea, but perhaps there's a more formal definition to this.
 

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