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Considering the Laurent series expansions of the function $$ f(z)=\frac{4z}{(z-1)(z-3)^2} $$ that are valid in the annulus: $ \{z\in{\mathbb C}:1<|z|<3\} $, we have $$ f(z)=\sum_{n=1}^{\infty}z^{-n}+\sum^{\infty}_{n=0}\frac{2n+3}{3^{n+1}}z^n\qquad\text{for}~1<|z|<3. $$ I am confused about what is $\operatorname{Res}(z_0,f)$ where $z_0=0$. From the formula above, it seems that $\operatorname{Res}(0,f)=1$. But on the other hand, $f$ is analytic at $x_0=0$, then $\operatorname{Res}(0,f)=0$.

  • What is the mistake I made here?
  • The definition of residue in complex analysis I learned is only for the meromorphic function $f$ at an isolated singularity $a$. Is the notation $\operatorname{Res}(z_0,f)$ meaningful for any point on the complex plane?

[ADDED]
BACKGROUND: I got this problem when I went over some calculation in complex analysis in the book Cracking the GRE Mathematics Subject Test(4th Edition). The author calculates the integral $$ \oint_C\frac{4z}{(z-1)(z-3)^2}\qquad C=\{z\in{\mathbb C}:|z|=2\}~~\text{counterclockwise} $$ as the following:

The Laurent series of the integrand is $$ f(z)=\sum_{n=1}^{\infty}z^{-n}+\sum^{\infty}_{n=0}\frac{2n+3}{3^{n+1}}z^n\qquad\text{for}~1<|z|<3. $$ Since the coefficient of the $z^{-1}$ term in this series is $a_{-1}=1$, we have $$ \oint_C\frac{4z}{(z-1)(z-3)^2}=2\pi i\cdot a_{-1}=2\pi i $$

However, what I learned is that from residue theorem, this calculation is supposed to be $$ \oint_C\frac{4z}{(z-1)(z-3)^2}=2\pi i\cdot \operatorname{Res}(1,f)=2\pi i $$ I am wondering if the author's calculation is another version of residue theorem. But the $a_{-1}$ he used is not $\operatorname{Res}(0,f)$ if $\operatorname{Res}(0,f)=0$ is the answer to the first question above.

  • The residue is meaningful anywhere in the open domain of the function $f(z)$, as $\operatorname{Res}{z_0}(f) = \oint{0} f(z_0 + z) \mathrm{d} z$. The residue of your particular function at $z_0 = 0$ is zero, because $f$ is analytic there. – Sasha Oct 14 '11 at 16:27
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    You noted yourself that the series converges and reprsents $f$ only in the annulus $1<|z|<3$. But the residue at the origin is about the behavior of your function near the origin. This series is not valid near the origin, so cannot be used to compute the residue there. – Jyrki Lahtonen Oct 14 '11 at 16:35
  • A related problem. – Mhenni Benghorbal Jan 25 '13 at 05:20

1 Answers1

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These are actually two different ways to calculate the integral.

1.The first is the one you quoted in the book:

Since $f$ is holomorphic in the annulus $\{z\in{\Bbb C:1<|z|<3}\}$, we have the Laurent series $$ f(z)=\sum_{n=1}^{\infty}z^{-n}+\sum^{\infty}_{n=0}\frac{2n+3}{3^{n+1}}z^n\qquad\text{for}~1<|z|<3. $$ Since the coefficient of the $z^{-1}$ term in this series is $a_{-1}=1$, we have $$ \oint_C\frac{4z}{(z-1)(z-3)^2}=2\pi i\cdot a_{-1}=2\pi i. $$

This is actually not from residue theorem. It is from another fact that the Laurent series converges in any compact subset of the annulus and thus one can do the integration term by term, i.e., one can exchange the operation of $\int$ and $\sum$. On the other hand, only the term $\int_{C}1/z dz$ is non-zero and this is where $2\pi i$ from.

2.The definition of residue of $f$ at $z_0$ res$_{z=z_0}f$ is defined as the coefficient of $1/(z-z_0)$ in the Laurent series in the deleted neighborhood of $z_0$, that is, the set $\{z:0<|z-z_0|<r\}$ for some $r>0$. When $f$ is holomorphic at $z_0$, the coefficient of $1/(z-z_0)$ in the Laurent series is $0$ (the whole principal part is $0$ actually). Hence, in your case, res$_{z=0}f=0$.

On the other hand, the residue theorem is stated as the following

Suppose that $f$ is holomorphic in an open set containing a circle $C$ and its interior, except for a pole at $z_0$ inside $C$. Then $$ \int_{C}f(z)dz=2\pi i \cdot{\text res}_{z=z_0}f. $$

Since $z=1$ is the pole inside $C$, the residue you need to find is res$_{z=1}f$ rather than res$_{z=0}f$.