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Cauchy's integral (therom aand formula)

  1. Apr 24, 2014 #1
    let D be simple connected Domain and C be simple close curve in D.

    then by cauchy integral theorm ∫(z) dz over C is zero.→let this be my 1st equation.

    but by cauchy integral formula for a point (a) inside C, we can say

    f(a)=(1/2∏i)[closed integral over c]∫f(z)dz/(z-a)

    NOTE f(z) is analytic function in D.

    but substituting [closed integral over c]∫f(z)dz=0 from equation 1,

    we get f(a)=0, for all a belonging to C but which is not true.

    now where i am wrong ?

    i don't get the intuition behind cauchy integral therom and formula ,

    would be glad if someone helped me.
  2. jcsd
  3. Apr 24, 2014 #2


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    It's 't-h-e-o-r-e-m'.
  4. Apr 24, 2014 #3


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    You mean [tex]\int f(z)dz[/tex] over C is zero when is analytic inside C? Of course [itex]\int z dz= 0[/itex] but that is irrelevant to what you have below.

    Apparently you forgot about the [tex]\frac{1}{z- a}[/tex] part! If f(z) is analytic inside C, [tex]\frac{f(z)}{z- a}[/tex] is not- it has a pole of order 1 at z= a.
  5. Apr 24, 2014 #4
    If you want intuition for this stuff, you should read Visual Complex Analysis. I could hardly do it justice here.
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