Characterizations of poles

Suppose $a$ is a pole of order $m$. Then [L4] gives a holomorphic extension $g$ of $(z-a{)}^{m}f(z)$ with $g(a)\mathrm{\ne }0$. The singularity of $g$ at $a$ is removable, so [L5] writes $g(z)={\sum }_{n\ge 0}{b}_n(z-a{)}^n$ near $a$ with $b_0=g(a)\mathrm{\ne }0$; dividing by $(z-a{)}^m$ gives $f(z)={\sum }_{n\ge 0}{b}_n(z-a{)}^{n-m}$, whose principal part is finite and nonzero and ends at $(z-a{)}^{-m}$.

Suppose the principal part is finite and nonzero, and let $m$ be the largest index with $c_{-m}\mathrm{\ne }0$. Then $g(z):=(z-a{)}^{m}f(z)=c_{-m}+{\sum }_{k\ge 1-m}{c}_k(z-a{)}^{k+m}$ has zero principal part, so [L1] makes $g$ holomorphic at $a$ with $g(a)=c_{-m}\mathrm{\ne }0$. Therefore $a$ is a pole of order $m$ by [L4].

Suppose $\mathrm{\mid }f(z)\mathrm{\mid }\to \mathrm{\infty }$ as $z\to a$. Then $f$ is nonzero on some punctured neighbourhood of $a$, so $h:=1\mathrm{/}f$ is holomorphic there by [L3], and $h(z)\to 0$. By [L1], $h$ extends holomorphically across $a$ with value $0$, proving condition 4.

Suppose condition 4 holds. By [L2], the extension of $1\mathrm{/}f$ factors as $(z-a{)}^{m}u(z)$ for some $m\ge 1$ and some holomorphic $u$ with $u(a)\mathrm{\ne }0$; shrinking the disc if needed, $u$ stays nonzero there, so $f(z)=(z-a{)}^{-m}u(z{)}^{-1}$ and $a$ is a pole of order $m$ by [L3] and [L4].

The extension $g$ of step 1.1 is continuous and nonzero at $a$, so $\mathrm{\mid }g(z)\mathrm{\mid }\ge \delta >0$ near $a$; therefore $\mathrm{\mid }f(z)\mathrm{\mid }=\mathrm{\mid }g(z)\mathrm{\mid }\mathrm{\mid }z-a{\mathrm{\mid }}^{-m}\ge \delta \mathrm{\mid }z-a{\mathrm{\mid }}^{-m}\to \mathrm{\infty }$.

Step 1.1 proves $1\Rightarrow 2$, step 2.1 proves $1\Rightarrow 3$, step 1.2 proves $2\Rightarrow 1$, step 1.3 proves $3\Rightarrow 4$, and step 1.4 proves $4\Rightarrow 1$; hence all four conditions are equivalent, and the pole order is the largest negative exponent present in the finite principal part.