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Indefinite Integrals in the Complex domain

  • Thread starter kathrynag
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  • #1
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Homework Statement





I'm looking at integrals about the unit circle and need to show that they equal 0.
integral[e^zdz/(z-2)]
integral(sinz/zdz)
integral[sinhz/(z^2+2z+2)dz]
My thoughts on showing these is using series. Is this the right idea?





Also I'm looking at a couple of integrals and will be evaluating along any curve joining the end points of integration.
integral(cosh3zdz) from -1 to i series expansion?
integral(z^3-iz)dz from -R to R This just confuses me with the R's. Is this just a normal integral, but converting z to x+iy?
 

Answers and Replies

  • #2
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You have to show that the functions do not have any singular points inside the circle. That's clear, except perhaps for the third case. So, equate the numerator z^2+2z+2 to zero and solve for the singular points.

The integrals along any curve can be done by finding the antiderivative and then subtracting the values of the antiderivative at the end point and starting pooint, just like you would evaluate an ordinary integral of a (real) function over a segment of the real axis.
 
  • #3
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Well, what about sinz/z? wouldn't 0 be a problem point?
 
  • #4
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Well, what about sinz/z? wouldn't 0 be a problem point?
The limit of sin(z)/z for z ---> 0 exists, it is equal to 1.

This means that you can re-define the function to be f(z) = sin(z)/z everywhere except at z = 0, at z = 0 you define the function to be 1.

Now, since you are integrating f(z) around the unit circle, this redefinition does not affect the outcome of the integral. But since f(z) defined in this way is an analytic function, the integral is zero.

If you have studied till Cauchy's theorem that says that a contour integral of a complex differentiable function is zero, then you should prove that f(z) is complex differentiable at z = 0. If you have studied beyond this and know that complex differentiable implies analiticity and vice versa then that is not necessary as it is obvious that sin(z)/z is analytic.
 

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