Poles, branch cuts, and zeros

Question

From what I understand, these three concepts all describe the points where the function is not continuous. How to tell them apart? Thanks!

Answer 1

• If $f(z)$ is holomorphic/analytic on $0 < |z-z_0| < r$ then $z_0$ is an isolated singularity. From the Cauchy integral formula in an annulus you have the Laurent series $f(z) = \sum_{n=-\infty}^\infty a_n (z-z_0)^n$ converging on $0 < |z-z_0| < r$, and two cases are possible :

  • $a_n = 0 $ for $n < -k$ so that $(z-z_0)^{k}f(z)$ is analytic on $|z-z_0| < r$ and $z=z_0$ is a pole of order $k$. If $k\le 0$ then $z_0$ was in fact a removable singularity

  • otherwise $z= z_0$ is an essential singularity of $f(z)$

• Other types of singularities are non-isolated and include :

  • branch points : a point around which you can continue analytically $f(z)$, but $f(z_0+e^{2i \pi}(z-z_0)) \ne f(z)$)

  • and frontiers : $f(z) = \sum_{n=1}^\infty z^{2^n}$ is analytic on $|z| < 1$ but $\lim_{r \to 1^-} f(r e^{2 \pi i m/2^k}) = \infty$ whenever $m,k \in \mathbb{N}$, so you can't continue analytically $f(z)$ beyond $|z| < 1$

Answer 2

$z_0$ is a pole of $f$ iff $f$ is analytic on $\{z: 0<|z-z_0|<r\}$ for some $r>0,$ and $f(z_0)$ cannot be defined in such a way that $f$ is analytic on $\{z:|z-z_0|<r\},$ but also that $f(z)=(z-z_0)^ng(z)$ for $0<|z-z_0|<r$ for some $n\in \mathbb N,$ where $g$ is analytic on $\{z:|z-z_0|<r\}.$ For example if $f(z)=1/z+2/(z-1)^3$ then $f$ has a pole at $0$ and at $1.$

$z_0$ is a removable singularity of $f$ iff $f$ is analytic on $\{z:0<|z-z_0|<r\}$ and either (i) $f(z_0)$ is not defined, or (ii) $f(z_0)$ is defined but $f$ is not continuous at $z_0;$ but there is a (unique) value that be be assigned (or re-assigned) as $f(z_0)$ so that $f$ is analytic on $\{z: |z-z_0|<r\}.$

$z_0$ is an essential singularity iff $f$ is analytic on $\{z:0<|z-z_0|<r\}$ and $z_0$ is not a pole or a removable singularity. For example if $f(z)=e^{1/z}$ for $z\ne 0$ then $0$ is an essential singularity of $f.$

A zero of $f$ is any $z$ such that $f(z)=0.$