28 Jul 2025
Singularities, Branch Points, and Branch Cuts
Removable singularities, poles, essential singularities, and the cuts that make multivalued functions single-valued.
A point $z_0$ is an isolated singularity of $f$ if $f$ is analytic in a punctured disk
\[0<\lvert z-z_0\rvert<R\]but is not analytic at $z_0$ itself.
Removable singularity
If $\lim_{z\to z_0}f(z)=L$ is finite, define $f(z_0)=L$. The singularity is then removed.
For
\[f(z)=\frac{z^2-1}{z-1},\]factorization gives, for $z\ne1$,
\[f(z)=\frac{(z-1)(z+1)}{z-1}=z+1.\]Therefore
\[\lim_{z\to1}f(z)=2.\]Setting $f(1)=2$ makes the function analytic at $z=1$.
Pole
The point $z_0$ is a pole of order $m$ if
\[\boxed{ \lim_{z\to z_0}(z-z_0)^mf(z)=A, \qquad 0<\lvert A\rvert<\infty},\]and no smaller positive integer has this property.
For
\[f(z)=\frac{1}{(z-z_0)^m},\]$z_0$ is a pole of order $m$. The magnitude diverges as $\lvert z-z_0\rvert^{-m}$.
Essential singularity
An isolated singularity that is neither removable nor a pole is essential. For
\[f(z)=e^{1/z},\]take $z=x>0$ and let $x\to0$: then $e^{1/x}\to\infty$. Along $z=x<0$, however, $e^{1/x}\to0$. The limit is not finite, so the singularity is not removable. For every fixed $m$,
\[z^me^{1/z}\to\infty\]along $z=x\to0^+$, so no finite pole order exists. Therefore $z=0$ is essential.
Branch points and branch cuts
Write
\[z=re^{i(\theta+2\pi k)}.\]Then
\[\sqrt z=\sqrt r\,e^{i(\theta+2\pi k)/2} =(-1)^k\sqrt r\,e^{i\theta/2}.\]After one circuit around the origin, $\theta\mapsto\theta+2\pi$ and $\sqrt z$ changes sign. The value does not return to itself, so $z=0$ is a branch point.
Similarly,
\[\log z=\ln r+i(\theta+2\pi k)\]changes by $2\pi i$ after one circuit. A branch cut removes a curve joining branch points or extending from a branch point to infinity, preventing such a circuit inside the chosen domain. For the principal branches, one common choice is the negative real axis, with
\[-\pi<\arg z<\pi.\]
The cut is a convention; the multivalued behavior at the branch point is intrinsic.
Discussion