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My definition of the theorem is stated in this section of the Wikipedia article.

My question is why point $a$ must be within the interior of $D$, and can't it also be in $U$ (where I mean the part of $U$ but not $D$ the domain where $f$ is analytic)(I think $D$ is smaller since $D$ is any simple closed contour lying entirely within $U$)?

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    By being in the interior of D it is already in U since $D\subset U$... what exactly are you asking?2011-07-17
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    i mean isn't the graph of u is bigger than D and point a could be in u but not in D ?2011-07-17
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    @ dylan moreland,what do you mean by a is in in U∖∂D2011-07-17
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    @dylan moreland-may you answer my question or give me some example in your example?2011-07-17
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    $U$ is a superset of $D$... yes. $f$ is holomorphic on the open set $U$. Suppose $a$ is a point within $U$, then you can form a tiny circle around $a$ contained still within $U$ so that you can apply the integral formula to evaluate $f(z)/(z-a)$, where the contour is exactly that circle within which $a$ is contained.2011-07-17
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    i am really confused-isn't the contour is just f(z)?2011-07-17
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    @Victor: I suggest you brush up on some of the prereqs of studying functions of a complex variable before trying to understand this theorem...2011-07-17

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If $a$ is not in $D$, then $f(z)/(z-a)$ is holomorphic in $D$. So the integral around the boundary of $D$ (the contour) is equal $0$ by Cauchy's Theorem.