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Let $D$ be closed unit disk in $\Bbb C $, and $C$ be the unit circle. Let $a\in\text{interior}(D)$.

For any continuous function $f : D -> \Bbb C$ We define a function $\displaystyle F(a)=\oint_C\frac {f(z)}{(z-a)} $

Ahlfors "complex analysis " says that $F(z)$ is always analytic inside interior($D$)

Observe that $\displaystyle F(a) = 0 $ for $z^n, n <0 $ .

Also observe that if $f$ be a meromorphic function on $D$ with only one singularity at $0$ . Then we easily get that $\displaystyle F(a)= $ principal part of the Laurent series of $f$ at $0$.

This is what I observed!

now main question

What will happen in general case, (i.e. if we have more than one singularity, maybe poles or essential). I agree that $ F(z) = f(a) + \sum\limits_{x=\text{singularity of }f} \mathop{Res}_x \frac {f(z)}{z-a} $ . Is there something to do with principle parts at singularities of f ? as was observed in single singularity case?

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rohit
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    What is $F$? If you're defining $F$ by that integral, then how could the "equality" not be true? – Ron Gordon May 04 '13 at 11:37
  • @rohit: "power series" usually means positive powers. You seem to be asking something about the Laurent series. I don't understand the question however. If your function $f(z)$ has singularities inside the contour of integration, then Cauchy integral formula is no longer valid, and the Cauchy integral is almost by definition given by $\displaystyle f(a)+\sum_k\mathrm{res}_{z_k}\frac{f(z)}{z-a}$, where the sum is taken over the singularities mentioned above. In your example, there is an accidental compensation of $f(a)$ by the pole at $0$, which is due to the absence of singularity at infinity. – Start wearing purple May 04 '13 at 11:50
  • We can calculate $ res_b\frac{f(z)}{z-a}$ as follows.consider laurent series at z=b $\sum _{-\infty}^{\infty} \frac {a_n}{(z-b)^n}$ multiply it by $ \frac {1}{z-a} = \frac {1}{b-a} (1 - \frac{z-a}{b-a} + \frac{(z-a^2}{(b-a)^2} + ... ) $ is it alright? – rohit May 04 '13 at 16:54

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Hint: Note that, you have two poles inside the contour $|z|=1$, namely, $z=0$ with order $n$ and $z=a$ with order $1$ (simple pole). See here for how to compute residues for poles of order $n$.

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Mhenni Benghorbal
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An idea (to develop by you): draw little, tiny circles around zero and $\,a\,$ (say, each circle of radius $\,\epsilon:=\frac{|a|}{3}\,$, and "join" the exterior circle $\,C\,$ with these two circles by means of two little straight segments of line, so that you go around $\,C\,$ , get into the first tiny circle, go out on the same segment (and in the opposite direction so the integral on these segments cancel each other...), continue on C, get into the second tiny circle...etc.

You're then left with the simple integrals

$$\oint\limits_{C(0)_\epsilon}\frac{\frac1{z-a}}{z^n}=\left.\frac{2\pi i}{(n-1)!}\frac d{dz^{n-1}}\left(\frac1{z-a}\right)\right|_{z=0}$$

$$\oint\limits_{C(a)_\epsilon}\frac{\frac1{z^n}}{z-a}dz=\left.2\pi i\left(\frac1{z^n}\right)\right|_{z=a}$$

evaluated by means of Cauchy's Integral Formula

DonAntonio
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