Winding number version of Cauchy integral formula: Difference between revisions

Revision as of 14:31, 19 April 2008

Statement

Suppose $U$ is an open subset of $C$ and $f : U \to C$ is a holomorphic function.

Suppose $c$ is a sum of oriented piecewise smooth loops $c_{1}, c_{2}, \dots, c_{r}$ , each completely inside $U$ . In other words, when we use the symbol $\oint_{c} g (z) d z$ , we actually mean $\sum_{1}^{r} \int_{c_{r}} g (z) d z$ .

Denote by $n (c; z_{0})$ the sum of the winding numbers of the $c_{k}$ about $z_{0}$ . Suppose $c$ is zero-homologous, i.e. $n (c; z) = 0$ for $z \in C ∖ U$ . Then we have, for any $z_{0} \in U$ :

$n (c; z_{0}) f (z_{0}) = \frac{1}{2 π i} \oint_{c} \frac{f (z)}{z - z_{0}} d z$

Proof

The idea here is to define analytic functions on two open subsets, show that they agree on the overlap, and hence obtain an analytic function on the whole of $C$ .

Define first:

$φ (z, w) = \frac{f (z) - f (w)}{z - w}$ if $z \neq w$ , $f^{'} (z)$ if $z = w$

Observe that $φ$ is analytic in each variable. To see this, note that for fixed $z$ , $φ$ is analytic in $w$ for $w$ away from $z$ , and the power series shows that it is analytic in a neighborhood of $z$ .

Now define an open subset:

$V : = {z \in C ∖ c ∣ n (c; z) = 0}$

Then $U \cup V = C$ , and $V$ is an open subset. Consider the function $g_{1}, g_{2} : U, V \to C$ :

$g_{1} (z) : = \oint_{c} φ (z, w) d w$

and:

$g_{2} (z) : = \oint_{c} \frac{f (w)}{w - z} d w$

To see that $g_{1}$ and $g_{2}$ agree on the intersection, observe that on the intersection $U \cap V$ , the difference is given by:

$\oint_{c} \frac{f (z)}{z - w} d z$

which is zero for points in the intersection, as they are points in $U$ about which the winding number is zero.

Thus, we can paste them together to get a single holomorphic function $g : C \to C$ . It is clear from the expression that $g (z) \to 0$ as $| z | \to \infty$ ; hence, using the fact that any bounded entire function is constant, we obtain that $g \equiv 0$ . This shows that:

$\oint_{c} \frac{f (z)}{z - w} d w = \oint_{c} \frac{f (w)}{z - w} d w$

The left side evaluates to $n (c; z) f (z)$ : precisely what we want.

@@ Line 25: / Line 25: @@
 Then <math>U \cup V = \mathbb{C}</math>, and <math>V</math> is an open subset. Consider the function <math>g_1, g_2:U, V \to \mathbb{C}</math>:
-<math>g_1(z) := \oint_c \varphi(z,w) \, dz</math>
+<math>g_1(z) := \oint_c \varphi(z,w) \, dw</math>
 and:
@@ Line 36: / Line 36: @@
 which is zero for points in the intersection, as they are points in <math>U</math> about which the winding number is zero.
+Thus, we can paste them together to get a single holomorphic function <math>g: \mathbb{C} \to \mathbb{C}</math>. It is clear from the expression that <math>g(z) \to 0</math> as <math>|z| \to \infty</math>; hence, using the fact that [[bounded and entire implies constant|any bounded entire function is constant]], we obtain that <math>g \equiv 0</math>. This shows that:
+<math>\oint_c \frac{f(z)}{z - w} \, dw = \oint_c \frac{f(w)}{z - w} \, dw</math>
+The left side evaluates to <math>n(c;z)f(z)</math>: precisely what we want.