Let $C_0$ denote the circle centered around some point $z_0\in\mathbb{C}$ with radius $R$. We can parametrize this circle like this:

$$\begin{array}{cc} z(\theta)=z_0+Re^{i\theta}, & \theta \in [-\pi,\pi]\end{array}$$

We are asked to show that:

$$\int_{C_0}(z-z_0)^{n-1}dz = \left\lbrace \begin{array}{ll} 0 & n\in\mathbb{Z}\backslash\{0\} \\ 2\pi i & n=0\end{array}\right. $$

I have completed the exercise but do not quite understand what this gives us. Could someone please provide some form of intuition for this?


2 Answers 2


What this gives is something remarkably powerful. Specifically, the result tells you that there's a way to isolate the various terms of a polynomial, and in fact a power series.

Think of it this way: take $z_0=0$. We then have that \begin{equation*} \frac{1}{2\pi i}\int_{C_0}{z^{n-1}}\,dz=\begin{cases}1&n=0,\\0&\text{otherwise}.\end{cases} \end{equation*}

Any time you can obtain the Kronecker delta, it gives you the chance to select a specific piece of your input. In this case, let $P$ be a polynomial: $P(x)=a_0+a_1x+a_2x^2+\ldots+a_nx^n$.

Consider the integral \begin{equation*} \frac{1}{2\pi i}\int_{C_0}{\frac{P(z)}{z^{k+1}}}\,dz. \end{equation*} Expanding this out, we have \begin{align*} \frac{1}{2\pi i}\int_{C_0}{\frac{P(z)}{z^{k+1}}}\,dz &=\frac{1}{2\pi i}\int_{C_0}{\left(a_0z^{-k-1}+a_1z^{(1-k)-1}+\cdots+a_nz^{(n-k)-1}\right)}\,dz\\ &=\frac{a_0}{2\pi i}\int_{C_0}{z^{(0-k)-1}}\,dz+\frac{a_1}{2\pi i}\int_{C_0}{z^{(1-k)-1}}\,dz+\cdots+\frac{a_n}{2\pi i}\int_{C_0}{z^{(n-k)-1}}\,dz. \end{align*} By the result you proved, this is extremely easy to compute. Each integral $\int_{C_0}{z^{(\ell-k)-1}}\,dz$ is $0$ unless $\ell=k$, so we get \begin{align*} \frac{1}{2\pi i}\int_{C_0}{\frac{P(z)}{z^{k+1}}}\,dz &=0+\cdots+0+a_k+0+\cdots+0\\ &=a_k. \end{align*} Therefore, given a polynomial function $P$, we can extract the coefficients of the polynomial by integrating $P(z)$ on a circle about $0$. This works for power series as well... and changing the center of the circle just moves us from a power series in $z$ to a power series in $(z-z_0)$. (That is, from a Maclaurin series to a Taylor series.)

In other words, if $f$ is an analytic function, we can compute its derivatives at $0$ by computing integrals! We just need to compensate for the factor of $k!$ that appears in the coefficients of the power series, obtaining \begin{equation*} f^{(k)}(0)=\frac{k!}{2\pi i}\int_{C_0}{\frac{f(z)}{z^k}}\,dz. \end{equation*} In fact, \begin{equation*} f^{(k)}(z_0)=\frac{k!}{2\pi i}\int_{C_0}{\frac{f(z)}{(z-z_0)^k}}\,dz, \end{equation*} if the circle is centered at $z_0$.

P.S.: It's actually sufficient for the circle to contain $z_0$, even if it has a different center. In fact, it doesn't even need to be circular! Explaining that leads to some truly beautiful mathematics... all the way to Cauchy's integral formula, and its many fascinating consequences.

  • $\begingroup$ Naturally, Spencer posted while I was still writing... we're both getting at the same ideas. $\endgroup$
    – Eric Astor
    Feb 24, 2014 at 4:10

Consider the function $f(z)=\sum_{-\infty}^\infty c_n z^n$. Your result tells us that if we want to determine the coefficient $c_{-1}$ we just need to integrate $f(z)$ over a closed contour about the origin.

$$ \int_C f(z) dz = 2\pi i c_{-1}$$

If we want some other coefficient we can multiply by some power of $z$ first and then integrate. For instance integrating $z^2f(z)$ will allow us to determine the coefficient $c_{-3}$.

All of this hinges upon the fact that the integral is nonzero only when $z$ is raised to the $-1$ power. Another way of writing this is,

$$ \int_C z^n z^m dz = \delta_{n+m,-1}2 \pi i \equiv \begin{cases} 2\pi i \text{ if } m+n=-1 \\ 0 \text{ otherwise }\end{cases} $$

this is what is usually called an orthogonality relation and extends to applications outside of complex variable theory. This property provides us with a method for determining the coefficients for the expansion of our function in terms of some family of orthogonal functions. Some examples would be the fourier series expansion, hermite polynomial expansion, and the spherical harmonic expansion.

  • 1
    $\begingroup$ I didn't mention it but this identity is also the key to residue calculus which turns the whole argument on its head. Rather than using the integral to determine the coefficient you use the coefficient to evaluate hard integrals. $\endgroup$
    – Spencer
    Feb 24, 2014 at 3:55

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