Consider a function f holomorphic everywhere in a domain $\Omega$ except at a point of singularity $z_0$. If the function f was say $\frac{sin(z)}{z^3}$ the Laurent series expansion at $z_0 = 0$ would have no $z^{-1}$ term, so would the residue of this function be $0$ or would it not exist?

If it is $0$, does this mean that the residue always exist for an isolated singularity? Furthermore, since the Taylor series expansion is just the Laurent series with no negative power terms, meaning the the coefficient of $z^{-1}$ is zero, can we say that the residue always exists for a point in the domain? Regardless of whether it is a singularity or not.

  • 2
    $\begingroup$ The residue is just a (complex) number. It can be of course equal to 0. $\endgroup$
    – user
    Commented Apr 11, 2018 at 9:11
  • $\begingroup$ You can take the residue to be 0 when there is no singularity but this has no theoretical significance. Residue is useful only when you have an isolated singularity. $\endgroup$ Commented Apr 11, 2018 at 9:13

1 Answer 1


Given a holomorphic function $f:U\to\mathbb{C}$ where $U:=\Omega\setminus\{z_0\}$ for $z_0\in\Omega$ open. Then the residue of $f$ at $z_0$ always exists.

Let $r>0$ s.t. $\overline{B_r(z_0)}\subseteq\Omega$. Since $f$ is holomorphic, it follows by Cauchy's integral theorem for homotopic curves: $$ \mathrm{Res}_{z_0}(f) := \lim_{\varepsilon\to0} \frac{1}{2\pi i} \oint_{\partial B_\varepsilon(z_0)} f(z) \,dz = \lim_{\varepsilon\to0} \frac{1}{2\pi i} \oint_{\partial B_r(z_0)} f(z) \,dz = \frac{1}{2\pi i} \oint_{\partial B_r(z_0)} f(z) \,dz $$

The set $\partial B_r(z_0)$ is compact and $f$ is continuous, therefore $f$ assumes a maximum value on $\partial B_r(z_0)$: $$ \left|\mathrm{Res}_{z_0}(f)\right| = \left|\frac{1}{2\pi i} \oint_{\partial B_r(z_0)} f(z) \,dz\right| \leq r\cdot \max_{z\in\partial B_r(z_0)} |f(z)| < \infty $$


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