Cauchy Integral formula

  • Thread starter bodensee9
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  • #1
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Hello:

I am wondering about the following:

Let C be the circle |z| = 3. the contour integral

g(w) = integral on C of (2*z^2-z-2/z-w)dz can be evaluated by cauchy's integral formula. I am wondering what happens if w is greater than 3.

Would you get this: f = 2*z^2-z-2. This is an entire function.

because the contour integral of an analytic function f around 2 simply connected contours is equal, so for a contour C2 that is greater than |z| = 3, the integral of f over that contour would be equal to the integral of f over contour C. So, this means that g(w) would still give you the value of the contour integral even though your contour covered a greater area than |z| = 3. So, this would mean that for w > 3, you can still use the Cauchy formula and get 2*pi*i*2*w^2-w-2? Thanks very much.
 

Answers and Replies

  • #2
Dick
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The integral around two contours are the same only if they enclose the same poles and wind in the same direction. If |w|>3 then its not in your circular contour. If another contour does enclose the pole then its value is not the same as the circular contour.
 
  • #3
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Cauchy's Formula

Hello,

but sorry, could you give me some hints as to how to do the problem then?
 
  • #4
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I guess what i mean is, why can't i draw a big circle that covers all |w| >3 and still use the same integral formula? Thanks.
 
  • #5
Dick
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Because you want the integral over the circle C, not the integral over the bigger contour. You'll have to divide the problem into two cases. |w|<3 and |w|>3.
 
  • #6
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Oh, okay. so would the value be zero then? since the function, if |w| >3, is analytic throughout C (|z| = 3), and the integral of an analytic function over a simply connected closed contour = 0 if f is analytic everywhere inside and on the boundaries of the contour? Thanks.
 
  • #7
Dick
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Yes, the value would be zero for |w|>3.
 

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