Let $D$ be a simply connected domain, bounded by a piecewise smooth curve $\gamma$, and let $\,f$ be analytic in a set $U$ that contains the closure of $D$ (ie $D$ and $\gamma$). Then for all $w \in D$,

$$\begin{aligned} f(w) = \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{z - w} \,dz \\ \end{aligned}$$

!

For example, what is the value of

$$\begin{aligned} \oint_{\left| z \right| = 2} \frac{z^2}{z - 1} \,dz \\ \end{aligned}$$

?

This is not an easy integral to evaluate directly, but with Cauchy's formula, the problem can be reduced into simple pattern matching ! Can you see why the value is $2 \pi i$ ? Now, how about

$$\begin{aligned} \oint_{\left| z \right| = 1} \frac{z^2}{z - 2} \,dz \\ \end{aligned}$$

?

Don't get tricked, and hopefully you can see the value is clearly zero.

An amazing consequence of this formula is that as long as $f$ is analytic in an open set $U$, then $f'$ is also analytic in $U$ ! Indeed,

$$\begin{aligned} f'(w) = \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{(z - w)^2} \,dz \\ \end{aligned}$$

And in general,

$$\begin{aligned} f^{(k)}(w) = \frac{k!}{2\pi i} \oint_\gamma \frac{f(z)}{(z - w)^{k+1}} \,dz \\ \end{aligned}$$

!

Doesn't this look déjà vu and perhaps somehow related to the coefficient of the $k$th term of a Taylor series expansion ?

For example, what is the value of

$$\begin{aligned} \oint_{\left| z \right| = 2\pi} \frac{z^2\sin z}{(z - \pi)^3} \,dz \\ \end{aligned}$$

? It becomes easy to realize it's exactly $-4\pi^2i$. The incredible Cauchy !

More at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 9:12 AM