Singularities and Classification

We use the Laurent expansion (assumed known) as the starting analytic tool, together with basic estimates on Laurent coefficients and elementary limit arguments. The proof proceeds case by case; within each case, the equivalences are proved as a cycle of implications, which is logically airtight and avoids redundancy.

Preliminary estimate. For $0<\rho<r$, the Laurent coefficient formula gives, by the standard $ML$-estimate on the circle $|z-a|=\rho$,

This will be used repeatedly to bound coefficients using boundedness of $f$.

(a) Removable singularity. (i) $\Rightarrow$ (ii): If $c_n=0$ for all $n<0$, the Laurent series reduces to the power series $\sum_{n\geq0}c_n(z-a)^n$, which converges on all of $D(a,r)$ (Laurent series converge on their annulus of definition, and here the "inner radius" of convergence for the non-negative part extends to include $a$ since there is no singular part forcing an inner boundary). Define $\tilde f(z)=\sum_{n\ge0}c_n(z-a)^n$ on $D(a,r)$; this is holomorphic there (power series are holomorphic on their disk of convergence) and agrees with $f$ on $D'(a,r)$. Thus $\tilde f$ is the desired holomorphic extension. (ii) $\Rightarrow$ (iii): If $f$ extends to a holomorphic $\tilde f$ on $D(a,r)$, then $\tilde f$ is continuous at $a$, hence bounded on some closed subdisk $\overline{D(a,\rho)}\subset D(a,r)$ by continuity on a compact set. In particular $f=\tilde f$ is bounded on $D'(a,\rho)$. (iii) $\Rightarrow$ (iv): Suppose $|f(z)|\le K$ on $D'(a,\rho)$. Then for $z\ne a$ with $|z-a|<\rho$,

(iv) $\Rightarrow$ (i): Suppose $\lim_{z\to a}(z-a)f(z)=0$. In particular $(z-a)f(z)$ is bounded near $a$, say $|(z-a)f(z)|\le K$ for $0<|z-a|<\rho'\le\rho$, so $|f(z)|\le K/|z-a|$ there, giving $M(\rho)\le K/\rho$ for all small $\rho$. By the preliminary estimate, for each fixed $n<0$,

Since $n<0$, we have $-n-1 \ge 0$; write $n=-k$ with $k\ge1$, so the bound is $K\rho^{k-1}$. If $k\ge 2$ this already $\to 0$ as $\rho\to0^+$, forcing $c_n=0$. For $k=1$ (i.e. $n=-1$) we need the sharper hypothesis (iv) itself: given $\varepsilon>0$, choose $\rho$ small enough that $|(z-a)f(z)|<\varepsilon$ for $|z-a|=\rho$; then $M(\rho)\cdot \rho <\varepsilon$ on that circle is not quite the right form, so redo the estimate directly: from $|(z-a)f(z)|\le \varepsilon(\rho)$ on $|z-a|=\rho$ with $\varepsilon(\rho)\to0$, we get $M(\rho)\le \varepsilon(\rho)/\rho$, hence

so $c_{-1}=0$. Combined with the $k\ge2$ case, all $c_n=0$ for $n<0$. This closes the cycle (i)$\Rightarrow$(ii)$\Rightarrow$(iii)$\Rightarrow$(iv)$\Rightarrow$(i), proving all four are equivalent. This equivalence is exactly Riemann's removable singularity theorem.

(b) Pole of order $m$. (i) $\Rightarrow$ (ii): If $c_{-m}\ne0$ and $c_n=0$ for $n<-m$, write

where $g(z):=\sum_{k=0}^\infty c_{k-m}(z-a)^k$ (reindexing $k=n+m\ge0$). This is a power series with the same radius of convergence as the original Laurent series' outer part, hence converges and is holomorphic on all of $D(a,r)$, and $g(a)=c_{-m}\ne0$. (ii) $\Rightarrow$ (iii): Since $g$ is holomorphic at $a$ with $g(a)\ne0$, continuity gives $g(z)\to g(a)\ne0$ as $z\to a$, and $g$ is bounded away from $0$ near $a$ (by continuity, $|g(z)|>\tfrac12|g(a)|>0$ on some punctured disk around $a$). Then

since the numerator is bounded below by a positive constant and the denominator $\to0$. Moreover,

so the limit $\lim_{z\to a}(z-a)^m f(z)$ exists, is finite and nonzero. For uniqueness of such $m$: if $j<m$ then $(z-a)^j f(z) = (z-a)^{j-m}g(z)\to\infty$ (since $j-m<0$ and $g(a)\ne0$), so the limit does not exist finitely; if $j>m$ then $(z-a)^jf(z) = (z-a)^{j-m}g(z)\to 0$, giving the wrong (zero) limit. So $m$ is indeed the unique exponent giving a finite nonzero limit. (iii) $\Rightarrow$ (i): Suppose $|f(z)|\to\infty$ as $z\to a$ and $m$ is the unique positive integer with $L:=\lim_{z\to a}(z-a)^mf(z)$ finite and nonzero. Since $(z-a)^mf(z)\to L\ne0$ finite, this function is bounded near $a$: there exist $\rho_0>0,K>0$ with $|(z-a)^m f(z)|\le K$ for $0<|z-a|<\rho_0$. Hence $|f(z)|\le K|z-a|^{-m}$, so $M(\rho)\le K\rho^{-m}$ for small $\rho$. By the preliminary estimate, for $n<-m$, writing $n=-m-j$ with $j\ge1$,

forcing $c_n=0$ for all $n<-m$. It remains to show $c_{-m}\ne0$. Since $(z-a)^mf(z)\to L\ne 0$ and (from the vanishing of $c_n$, $n<-m$, just shown) $(z-a)^m f(z) = \sum_{k\ge0}c_{k-m}(z-a)^k$ is itself a convergent power series near $a$ whose value at $z=a$ (by continuity of power series) equals its constant term $c_{-m}$; matching with the limit $L$ gives $c_{-m}=L\ne0$. This closes the cycle (i)$\Rightarrow$(ii)$\Rightarrow$(iii)$\Rightarrow$(i).

(c) Essential singularity. (i) $\Leftrightarrow$ (ii): By definition, the singularity is removable iff all $c_n=0$ ($n<0$), a pole iff finitely many (but at least one) $c_n\ne0$ for $n<0$ terminating at some $-m$. Since the three cases (all zero / eventually zero, not all / not eventually zero) exhaust and are mutually exclusive for the coefficient sequence $(c_n)_{n<0}$, "essential" (infinitely many nonzero $c_n$, $n<0$) is logically identical to "neither removable nor a pole." This is immediate from the definition and requires no further analytic input. (ii) $\Rightarrow$ (iii) (Casorati–Weierstrass): Suppose, for contradiction, that for some $\rho\in(0,r)$ the image $f(D'(a,\rho))$ is not dense in $\mathbb{C}$. Then there exists $w_0\in\mathbb{C}$ and $\delta>0$ such that

Define $h(z) := \dfrac{1}{f(z)-w_0}$ on $D'(a,\rho)$. Then $h$ is holomorphic there (as $f-w_0\ne0$ and holomorphic), and $|h(z)| \le 1/\delta$ for all $z\in D'(a,\rho)$, so $h$ is bounded on a punctured disk around $a$. By part (a) [the equivalence (iii)$\Rightarrow$(i) already proved, i.e. Riemann's theorem], $h$ has a removable singularity at $a$; let $\tilde h$ be its holomorphic extension to $D(a,\rho)$. Case 1: $\tilde h(a) \ne 0$. Then near $a$, $f(z) = w_0 + \dfrac{1}{h(z)} = w_0 + \dfrac{1}{\tilde h(z)}$, and since $\tilde h$ is holomorphic and nonvanishing near $a$ (by continuity, as $\tilde h(a)\ne0$), the right side is holomorphic near $a$. So $f$ extends holomorphically to $a$, i.e. $f$ has a removable singularity — contradicting (ii). Case 2: $\tilde h(a) = 0$. Since $\tilde h$ is holomorphic and not identically zero near $a$ (it is nonzero on the punctured disk, as $h=1/(f-w_0)$ never vanishes there), the zero at $a$ has some finite order $k\ge1$: $\tilde h(z) = (z-a)^k u(z)$ with $u$ holomorphic and $u(a)\ne0$ near $a$ (this is the standard local factorization of zeros of a nonconstant holomorphic function, itself following from the Taylor expansion of $\tilde h$ at $a$). Then near $a$,

so $(z-a)^k(f(z)-w_0) = 1/u(z) \to 1/u(a) \ne 0$, finite. This is exactly criterion (iii) of part (b) (applied to $f - w_0$, which has the same singularity type as $f$ since $w_0$ is a constant shift and does not change the principal part), showing $f$ has a pole of order $k$ at $a$ — again contradicting (ii). Both cases contradict the hypothesis that $f$ is neither removable nor a pole at $a$. Hence the assumption that $f(D'(a,\rho))$ fails to be dense must be false, for every $\rho \in (0,r)$. This proves (iii). (iii) $\Rightarrow$ (ii): We prove the contrapositive. Suppose $f$ is removable or a pole at $a$. If removable, $f$ extends continuously (indeed holomorphically) to $a$, so $f(z)\to f(a)$ as $z\to a$; then for $\rho$ small, $f(D'(a,\rho))$ lies in a small disk around $f(a)$ and is certainly not dense in $\mathbb{C}$. If $f$ has a pole at $a$, by part (b) $|f(z)|\to\infty$ as $z\to a$, so for $\rho$ small, $f(D'(a,\rho))$ lies outside a large disk $\{|w|\le K\}$ and again is not dense in $\mathbb{C}$ (it misses the entire disk $\{|w|<K\}$). Either way (iii) fails for small $\rho$. This proves the contrapositive, hence (iii)$\Rightarrow$(ii). This closes the logical cycle for part (c) and completes the proof of the theorem. $\blacksquare$