Cauchy's Integral Formula.

Recall that if $ z_0 \in \mathbb{C}$ and $ R\geq 0$ , $ B(z_0,R)=\{ z \in \mathbb{C}; \vert z-z_0\vert < R \}$ (see def open ball). As we work in the plane, it can happen that this open ball is also called an open disk.

Theorem 5.4.1   Take a function $ f$ analytic in the disk $ D(z_0,R)$ , a positive number $ \rho$ such that $ 0 < \rho < R$ and a point $ \alpha$ such that $ \vert\alpha -z_0\vert < \rho$ . Then:

$\displaystyle f( \alpha )= \frac {1}{2 \pi i} \int_C \frac {f(z)}{z - \alpha} \; dz$    

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

Another formulation of the above formula is as follows:

$\displaystyle \int_C \frac {f(z)}{z- \alpha} \; dz = 2 \pi i \cdot f(z) \vert _{z=\alpha}$    

Example 5.4.2  

$\displaystyle \int_{\vert z\vert=2} \frac {z^2}{z-i} \; dz = 2 \pi i \cdot z^2\vert _{z=i} = 2 \pi i \cdot (-1) = - 2 \pi i.$    

Example 5.4.3   We wish to compute the integral $ I= \int_{\vert z\vert=2} \frac {1}{z^2+1} \;
dz$ .

The denominator vanishes at two points, $ i$ and $ -i$ , both inside the contour. We will decompose this contour into two Jordan curves by the following way: draw the diameter of the circle which coincides with the x-axis. Denote:

Then $ I= \oint_{\vert z\vert=2} \frac {1}{z^2+1} \; dz = \oint_{C_1} \frac
{1}{z^2+1} \; dz + \oint_{C_2} \frac {1}{z^2+1} \; dz $ (the variable $ z$ ``travels'' twice on the diameter, but in opposite directions).

Figure 5:
\begin{figure}\mbox{\epsfig{file=twohalfcircles.eps,height=3cm}}\end{figure}

Therefore we have:

$\displaystyle I$ $\displaystyle = \int_{\vert z\vert=2} \frac {1}{z^2+1} \; dz$    
$\displaystyle \quad$ $\displaystyle = \int_{\vert z\vert=2}=\frac {1}{(z-i)(z+i)} \; dz$    
$\displaystyle \quad$ $\displaystyle = \int_{C_1} \frac {\frac {1}{z+i}}{z-i} \; dz + \int_{C_2} \frac {\frac {1}{z-i}}{z+i} \; dz$    
$\displaystyle \quad$ $\displaystyle = 2 \pi i \cdot \left[ \frac {1}{z+i} \right]_{z=i} + 2 \pi i \cdot \left[ \frac {1}{z-i} \right]_{z=-i}$    
$\displaystyle \quad$ $\displaystyle = 2 \pi i \cdot \left( \frac {1}{2i} - \frac {1}{2i} \right) =0.$    

Noah Dana-Picard 2007-12-24