Generalization of Cauchy's Integral Formula.

Notation: If $z_0 \in \mathbb{C}$ and $R\geq 0$ , $D(z_0,R)=\{ z \in \mathbb{C}; \vert z-z_0\vert < R \}$ .

Theorem 5.5.1 Take a function analytic in the disk , a positive number $\rho$ such that $0 < \rho < R$ and a point $\alpha$ such that $\vert\alpha -z_0\vert < \rho$ . Then:

$\displaystyle f^{(n)}( \alpha )= \frac {1}{2 \pi i} \oint_C \frac {f(z)}{(z - \alpha)^{n+1}} \; dz$

where represents the circle whose center is $\alpha$ and radius is equal to $\rho$ .

Example 5.5.2 Let $\mathcal{C}= \{ z \in \mathbb{C}; \; \vert z\vert=2 \}$ (i.e. $\mathcal{C}$ is the centrer whose center is the origin and whose radius is equal to 2. Then:

$\displaystyle \oint_{\mathcal{C}} \frac {\sin z}{(z-1)^2} \; dz = 2 \pi i \; \left[ \frac {d}{dz} (\sin z) \right]_{z=1} = 2 \pi i \cos 1.$

Recall that the point 1 is a point interior to $\mathcal{C}$ .

Example 5.5.3 Let $\mathcal{C}= \{ z \in \mathbb{C}; \; \vert z\vert=3 \}$ . we wish to compute the following integral:

$\displaystyle I=\oint_{\mathcal{C}} \frac {e^z}{(z-1)(z+2)^3} \; dz.$

The two points

and

are interior to $\mathcal{C}$ . We decompose the given fraction into simple fractions, by the same method used in Calculus; we have:

$\displaystyle \forall z \in \mathbb{C} - \{ 1, -2 \}, \; \frac {1}{(z-1)(z+2)^3}=\frac {-1}{27(z+2)}-\frac {1}{9(z+2)^2}-\frac {1}{3(z+2)^3}+\frac {1}{27(z-1)},$

hence:

$\displaystyle \forall z \in \mathbb{C} - \{ 1, -2 \}, \; \frac {e^z}{(z-1)(z+2)... ...e^z}{27(z+2)}-\frac {e^z}{9(z+2)^2}-\frac {e^z}{3(z+2)^3}+\frac {e^z}{27(z-1)}.$

By additivity of the integral, we have:

$\displaystyle I$	$\displaystyle = - \oint_{\mathcal{C}} \frac {e^z}{27(z+2)} \; dz - \oint_{\math... ...}} \frac {e^z}{3(z+2)^3} \; dz + \oint_{\mathcal{C}} \frac {e^z}{27(z-1)} \; dz$
$\displaystyle \quad$	$\displaystyle = 2\pi i \left( - \frac {1}{27} [e^z]_{z=-2} - \frac 19 \left[ \f... ...eft[ \frac {d^2}{dz^2} (e^z) \right]_{z=-2} + \frac {1}{27} [e^z]_{z=1} \right)$
$\displaystyle \quad$	$\displaystyle = 2\pi i \left( - \frac {1}{27} e^{-2} - \frac 19 e^{-2} - \frac 13 e^{-2} + \frac {1}{27} e \right)$
$\displaystyle \quad$	$\displaystyle = 2\pi i \left( - \frac {13}{27} e^{-2}+\frac {1}{27} e \right).$