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