Zero and pole

For a holomorphic function that is not identically zero on a connected region, every zero has a finite order: the Taylor series at \(z_0\) starts at \(a_m(z-z_0)^m\) with \(a_m\neq0\). In practice the order is found by differentiating: it is the first \(m\) with \(f^{(m)}(z_0)\neq0\). Zeros are isolated: a zero never accumulates with other zeros of the same nonzero function, and the identity principle follows from this.

A pole of order \(m\) is an isolated singularity whose Laurent principal part ends at \(\frac{a_{-m}}{(z-z_0)^m}\) with \(a_{-m}\neq0\); equivalently, \(f(z)=\frac{h(z)}{(z-z_0)^m}\) with \(h\) holomorphic and \(h(z_0)\neq0\). A pole of order \(1\) is called simple. Near a pole, \(|f(z)|\to\infty\) along every path.

Orders add and subtract like exponents. If \(f\) and \(g\) have zeros of orders \(m\) and \(n\), the product has a zero of order \(m+n\). The quotient \(f/g\) has a zero of order \(m-n\) if \(m>n\), a finite nonzero value if \(m=n\), and a pole of order \(n-m\) if \(n>m\). For instance, \(\frac{1}{\sin^3 z}\) has a pole of order \(3\) at \(0\) because \(\sin z\) has a simple zero there.

A typical mistake is to read the order off the apparent exponent. In \(\frac{z^2-1}{(z-1)^3}\) the exponent \(3\) is misleading, because the numerator also vanishes at \(1\) and the pole has order \(2\). The safe rule is to factor out \((z-z_0)^m\) and check that what remains does not vanish. This notion lets us compute residues at poles without expanding the whole series.