Residue of complex function

Question

I know that if function is meromorphic then it will have resiue. My simple question is if function is not meromorphic then can it have a residue, because I am not getting any such statement in my book. If it is correct or incorrect then plz help me with suitable example.

Answer 1

Yes, if $f(z)$ is holomorphic/analytic on $0 < |z| < r$ then it has a Laurent series $$f(z) = \sum_{n=-\infty}^\infty a_n z^n$$ converging uniformly/absolutely on $\epsilon < |z| < r - \epsilon$, and for $0 < R < r$ : $$Res(f(z),0) = a_{-1} = \frac{1}{2i\pi} \int_{|z| = R} f(z)dz$$ The proof is with the Cauchy integral theorem and formula for holomorphic functions on an annulus $$f(s) = \frac{1}{2i\pi} \int_{|z| = B} \frac{f(z)}{z-s}dz- \frac{1}{2i\pi} \int_{|z| = A} \frac{f(z)}{z-s}dz$$ whenever $0 < A < |s| < B < r$

---

• Example $$f(z) = e^{1/z} = \sum_{n=-\infty}^0 \frac{z^n}{n!}$$ it is analytic on $0 < |z|$ i.e. $z=0$ is an isolated singularity, and $Res(e^{1/z},0) = 1$

• Counter-example $$g(z) = \log(z), \qquad h(z) = z^{1/2}$$ It doesn't have a residue at $z=0$ because $z=0$ isn't an isolated singularity : $\log(z),z^{1/2}$ aren't holomorphic on $0 < |z| < r$, they are not even continuous at the branch cut.