Laurent Series

Let there be two circular contours $C_2$ and $C_1$, with the radius of $C_1$ larger than that of $C_2$. Let $z_0$ be interior to $C_1$ and $C_2$, and $z$ be between $C_1$ and $C_2$. Now create a cut line $C_c$ between $C_1$ and $C_2$, and integrate around the path $C\equiv C_1+C_c-C_2-C_c$, so that the plus and minus contributions of $C_c$ cancel one another, as illustrated above. From the Cauchy Integral Formula,

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

$$= {1\over 2\pi i}\int_{C_1}{f(z')\over z'-z}\,dz'+ {1\over 2\pi i}\int_{C_c}{f(z')\over z'-z}\,dz'$$

$$\phantom{=} - {1\over 2\pi i}\int_{C_2}{f(z')\over z'-z}\,dz'- {1\over 2\pi i}\int_{C_c}{f(z')\over z'-z}\,dz'$$

$$= {1\over 2\pi i}\int_{C_1}{f(z')\over z'-z}\,dz'- {1\over 2\pi i}\int_{C_2}{f(z')\over z'-z}\,dz'.$$ (1)

Now, since contributions from the cut line in opposite directions cancel out,

$$f(z) = {1\over 2\pi i}\int_{C_1}{f(z')\over (z'-z_0)-(z-z_0)}dz' - {1\over 2\pi i}\int_{C_2}{f(z')\over (z'-z_0)-(z-z_0)}dz'$$

$$= {1\over 2\pi i}\int_{C_1}{f(z')\over (z'-z_0)\left({1 - {z-z_0\ov... ...\pi i}\int_{C_2}{f(z')\over (z-z_0)\left({{z'-z_0\over z-z_0}- 1}\right)}\, dz'$$

$$= {1\over 2\pi i}\int_{C_1}{f(z')\over (z'-z_0)\left({1 - {z-z_0\ov... ...i i}\int_{C_2}{f(z')\over (z-z_0)\left({1 - {z'-z_0\over z-z_0}}\right)}\, dz'.$$ (2)

For the first integral, $\vert z'-z_0\vert > \vert z-z_0\vert$. For the second, $\vert z'-z_0\vert < \vert z-z_0\vert$. Now use the Taylor Expansion (valid for $\vert t\vert < 1$)

$${1\over 1-t}= \sum_{n=0}^\infty t^n$$ (3)

to obtain

$$f(z) = {1\over 2\pi i}\left[{\int_{C_1}{f(z')\over z'-z_0}\sum_{n=0}^\in... ...')\over z-z_0}\sum_{n=0}^\infty \left({z'-z_0\over z-z_0}\right)^n\,dz'}\right]$$

$$= {1\over 2\pi i}\sum_{n=0}^\infty (z-z_0)^n\int_{C_1}{f(z')\over (z'-z_0)^{n+1}}\, dz'$$

$$\phantom{=} + {1\over 2\pi i}\sum_{n=0}^\infty (z-z_0)^{-n-1}\int_{C_2}(z'-z_0)^nf(z')\,dz'$$

$$= {1\over 2\pi i}\sum_{n=0}^\infty (z-z_0)^n\int_{C_1}{f(z')\over (z'-z_0)^{n+1}}\, dz'$$

$$\phantom{=} + {1\over 2\pi i}\sum_{n=1}^\infty (z-z_0)^{-n}\int_{C_2} (z'-z_0)^{n+1}f(z')\,dz',$$ (4)

where the second term has been re-indexed. Re-indexing again,

$$f(z) = {1\over 2\pi i} \sum_{n=0}^\infty (z-z_0)^n\int_{C_1}... ...fty }^{-1}(z-z_0)^n\int_{C_2}{f(z')\over (z'-z_0)^{n+1}}\,dz'.$$ (5)

Now, use the Cauchy Integral Theorem, which requires that any Contour Integral of a function which encloses no Poles has value 0. But $1/(z'-z_0)^{n+1}$ is never singular inside $C_2$ for $n\geq 0$, and $1/(z'-z_0)^{n+1}$ is never singular inside $C_1$ for $n\leq -1$. Similarly, there are no Poles in the closed cut $C_c-C_c$. We can therefore replace $C_1$ and $C_2$ in the above integrals by $C$ without altering their values, so

$$f(z) = {1\over 2\pi i}\sum_{n=0}^\infty (z-z_0)^n\int_{C}{f(z')\over (z'-z_0)^{n+1}}\, dz'$$

$$\phantom{=} + {1\over 2\pi i}\sum_{n=-\infty }^{-1}(z-z_0)^n\int_{C}{f(z')\over (z'-z_0)^{n+1}}\,dz'$$

$$= {1\over 2\pi i}\sum_{n=-\infty }^\infty (z-z_0)^n \int_C {f(z')\over (z'-z_0)^{n+1}}\,dz'$$

$$\equiv \sum_{n=-\infty }^\infty a_n(z-z_0)^n.$$ (6)

The only requirement on $C$ is that it encloses $z$, so we are free to choose any contour $\gamma$ that does so. The Residues $a_n$ are therefore defined by

$$a_n \equiv {1\over 2\pi i}\int_\gamma {f(z')\over (z'-z_0)^{n+1}}\, dz'.$$ (7)

See also Maclaurin Series, Residue (Complex Analysis), Taylor Series

References

Arfken, G. ``Laurent Expansion.'' §6.5 in Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, pp. 376-384, 1985.

Morse, P. M. and Feshbach, H. ``Derivatives of Analytic Functions, Taylor and Laurent Series.'' §4.3 in Methods of Theoretical Physics, Part I. New York: McGraw-Hill, pp. 374-398, 1953.