Exercises: Singularities

• False: \(f(z) \coloneqq 1/z, g(z) \coloneqq-1/z \implies f+g = 0\).
• False: \(f(z) = g(z) = 1/z \implies fg = 1/z^2\).
• True: write \(f(z) = \sum_{k\in {\mathbf{Z}}} c_k (z-z_0)^k\), which has infinitely many negative coefficients, and \(g(z) = \sum_{k\geq -N}d_k (z-z_0)^k\). Then \begin{align*} f(z) + g(z) = \sum_{k\leq -N-1}c_k(z-z_0)^k + \sum_{k\geq -N} (c_k + d_k)(z-z_0)^k ,\end{align*} which again has infinitely many negative coefficients.
• True: check the Laurent expansion directly: \begin{align*} \qty{ \sum_{k\geq -N} c_k (z-z_0)_k }^2 &= {c_{-N}(z-z_0)^{-N} + { \mathsf{O}}((z-z_0)^{-N+1})}^2 \\ &= (c_{-N})^2 (z-z_0)^{-2N} + { \mathsf{O}}((z-z_0)^{-2N+1}) .\end{align*} An easier alternative, use theorem 1.2 from S&S: write \(f(z) = (z-z_0)^{-N} g(z)\) where \(g\) is holomorphic and (importantly) nonvanishing in a neighborhood of \(z_0\). Then \(f(z)^2 = (z-z_0)^{-2N}(g(z))^2\), where \(g^2\) is again nonvanishing in a neighborhood of \(z_0\) since \({\mathbf{C}}\) is an integral domain.

Write \(f(z) = p(z)/q(z)\) and factor \(p\): a principal root is \(\omega = e^{i\pi 3}\), so \begin{align*} p(z) &= (z-\omega\zeta_3^0)(z-\omega\zeta_3^1)(z-\omega\zeta_3^2) \\ &= (z-e^{i\pi\over 3})(z-e^{3i\pi \over 3})(z-e^{5i\pi \over 3}) \\ &= (z+1)(z-\omega)(z-\overline{\omega}) ,\end{align*} so \(z=-1\) is a removable singularity of \(f\). Alternatively, note that \({z^3+1 \over z+1} = z^2-z+1\) and cancel the common term.

Note that \(z=0\) is a zero of order \(n=2\) of \(1/f(z)\), since \(1/f(z) = z^2h(z)\) where \(h\) is nonvanishing in a neighborhood of \(0\). Thus \(z=0\) is a pole of order \(n=2\) of \(f\). The residue is computed as \begin{align*} \mathop{\mathrm{Res}}_{z=0} f(z) &= {1\over (1-1)!} \lim_{z\to 0} {\frac{\partial }{\partial z}\,} (z-0)^2f(z) \\ &= {\frac{\partial }{\partial z}\,} {z^3+1\over z+1}\Big|_{z=0} \\ &= \qty{ {3z^2\over z+1} - {(z^3+1)\cdot 1\over (z+1)^2} }\Big|_{z=0} \\ &= -1 .\end{align*}

Alternatively, expand as a Laurent series about \(z=0\): \begin{align*} f(z) &= z^{-2}(z^3 + 1) {1\over 1+z} \\ &= (z + z^{-2})\sum_{k\geq 0}(-z)^k \\ &= (z-z^2 + z^3 - \cdots) + (z^{-2} - z^{-1}+ 1 - z + \cdots) ,\end{align*} and read off the coefficient of \(z^{-1}\).

\(f_1\): removable, evident from Laurent expansion at \(z=0\): \begin{align*} z^{-2}\operatorname{Log}(1+z)\sin(z) &= z^{-1}\qty{ \sum_{k\geq 1} { (-z)^k \over k} } \qty{z - { z^3 \over 3!} + { z^5 \over 5!} - \cdots} \\ &= z^{-2} \qty{ -z+ { z^2 \over 2} - { z^3 \over 3} + \cdots} \qty{z - { z^3 \over 3!} + { z^5 \over 5!} - \cdots} \\ &= z^{-2}\qty{ z^2(-1) + z^3\qty{1\over 2} + z^2\qty{ -{ 1 \over 3!} - { 1 \over 3} } + \cdots } \\ &= 1 + {z \over 2} + \cdots .\end{align*}

\(f_2\): essential, evident from a sequence like \(z_k \coloneqq\qty{k\cdot {\pi\over 2} }^{-1}\) which makes \(\sin(z_k)\) oscillate between 0 and 1.

\(f_3\): pole of order 1 with residue 1, evident after some slightly clever Laurent manipulations: \begin{align*} {1\over e^z-1} &= {1 \over z + {1\over 2}z^2 + \cdots} \\ &= {1 \over z\qty{1 + {1\over 2}z + \cdots} } \\ &\coloneqq{1\over z \qty{1 + p(z)}} && p(z) \coloneqq{1\over 2} z + {1\over 3!}z^2 + \cdots \\ &= z^{-1}\sum_{k\geq 0}(-p(z))^k z^k \\ &= z^{-1}\qty{1 - zp(z) + z^2p(z)^2 - z^3p(z)^3 + { \mathsf{O}}(z^4)} \\ &= z^{-1}\qty{1- { \mathsf{O}}(z^2) + { \mathsf{O}}(z^4) - { \mathsf{O}}(z^6) } \\ &={1\over z} - { \mathsf{O}}(z) + { \mathsf{O}}(z^3) .\end{align*}

It suffices to show that if \(f(a) = 0\) then \(f\) is nonzero on some \({\mathbb{D}}_{\varepsilon}^*(a)\). Without loss of generality, suppose \(a=0\) and expand \(f(z) = \sum_{k\geq 0}c_k z^k = \sum_{k\geq m}c_k z^k\) where \(m\geq 0\) is minimal such that \(c_m\neq 0\). This exists since \(f\) is not identically zero, by uniqueness of power series. Then write \begin{align*} f(z) = \sum_{k\geq m} c_k z^k = z^m \sum_{k\geq m} c_k z^{k-m} = z^m (c_m + c_{m+1}z + \cdots) \coloneqq z^m g(z) ,\end{align*} where \(g(a) = c_m \neq 0\). Being nonzero is an open condition, so \(g\) is nonzero in some punctured neighborhood of \(a\), making \(f\) nonzero there.

Write \begin{align*} f(z) = \sin(\pi z) = (2i)^{-1}(e^{i\pi z} - e^{-i\pi z}) = 0 \iff e^{i 2\pi z} = 1 = e^{i 2k\pi} \iff 2\pi z = 2k\pi \iff z=k\in {\mathbf{Z}} .\end{align*} To see that these zeros are order one, write \begin{align*} \sin(\pi z) &= \sin(\pi(z-k) + k\pi) \\ &= \pm \sin(\pi(z-k)) \\ &= \pm\qty{ \pi(z-k) - {\pi^3\over 3!}(z-k)^3 + \cdots } \\ &= (z-k)^1 \cdot \pm \qty{ \pi - {\pi^3\over 3!}(z-k)^2 + \cdots } \coloneqq(z-k)g(z) \\ \end{align*} where \(g(k) = \pm \pi \neq 0\), making \(z=k\) an order 1 zero.

For the residues: \begin{align*} \mathop{\mathrm{Res}}_{z=k} \csc(\pi z) = \lim_{z\to k} (z-k)\csc(\pi z) { \overset{\scriptscriptstyle\text{LH}}{=} }\sec(k\pi) = (-1)^{k+1} .\end{align*}

• \(z=0\) of order 3: if \(z_0\) is order \(n\) for \(f\), then it’s order \(kn\) for \(f^k\). So check that \(e^z-1\) has a root \(z=0\) and \({\frac{\partial }{\partial z}\,}e^z-1\mathrel{\Big|}_{z=0} = e^z\mathrel{\Big|}_{z=0}\neq 0\), making it order 1.

• Order 2: \begin{align*} \lim_{z\to 0}z^0 f(z) &= \infty\\ \lim_{z\to 0}z^1 f(z) &= \infty\\ \lim_{z\to 0}z^2 f(z) &= 1 \neq 0 ,\end{align*} using that \(z/\sin(z) \overset{z\to 0}\longrightarrow 1\).

• Order 2: apply L’Hopital as necessary \begin{align*} \lim_{z\to 0}z^0 f(z) &= \infty\\ \lim_{z\to 0}z^1 f(z) &= \infty\\ \lim_{z\to 0}z^2 f(z) &= 1 \neq 0 .\end{align*} Alternatively, take the Laurent expansion: \begin{align*} f(z) &= z^{-4}(1 + z^2 + {1\over 2!} (z^2)^2 + {1\over 3!} (z^2)^3 + { \mathsf{O}}(z^4) - 1) \\ &= z^{-2} + {1\over 2!} + {1\over 3!} z^2 + { \mathsf{O}}(z^4) .\end{align*}

Without loss of generality suppose \(z_0=0\) is the pole. Write \(f(z) = \sum_{k\geq -N} c_k z^k\), then \begin{align*} f(z) = \sum_{1\leq j \leq N} c_j z^{-j} + \sum_{k\geq 0} c_k z^k \\ \implies f'(z) = \sum_{2 \leq j \leq N+1} -j c_j z^{-j-1} + \sum_{k\geq 1}k c_k z^{k-1} ,\end{align*} making \(0\) a pole of order \(N+1\).

Write the period lattice of \(f\) as \(\Lambda = \omega_1{\mathbf{Z}}+ \omega_2 {\mathbf{Z}}\), and without loss of generality (by translating \(P\) if necessary), assume that \(f\) has no poles on \({{\partial}}P\). Since \(P\) is bounded and \(f\) is periodic, if \(f\) has no poles then its only singularities will be removable. In this case \(f\), extends to a holomorphic function on \(P\), and thus an entire function, making \(f\) constant by Liouville. So \(f\) has at least one pole. Toward a contradiction, suppose \(f\) has exactly one pole \(z_0\in P\), in which case \(\int_{{{\partial}}P} f \neq 0\) since the residue at \(z_0\) will be nonzero. We’ll show that \(\int_{{{\partial}}P} f\) is forced to be zero to derive the contradiction.

Write \({{\partial}}P = \sum_{1\leq k \leq 4} \gamma_k\) where the \(\gamma_k\) are the edges traversed counterclockwise. By periodicity,

• \(I_1 \coloneqq\int_{\gamma_1} f = - \int_{\gamma_3}f\)
• \(I_2 \coloneqq\int_{\gamma_2} f = - \int_{\gamma_4}f\)

Thus \begin{align*} \int_{{{\partial}}P} f = \sum_{1\leq k \leq 4} \int_{\gamma_k} f = I_1 + I_2 - I_1 - I_2 = 0 .\end{align*} \(\contradiction\)

Note that if there are at least two poles, the residues may cancel and \(\int_{{{\partial}}P} f\) may be zero or nonzero. This argument in fact shows that the residues can not cancel, i.e. \(\sum_{k} \mathop{\mathrm{Res}}_{z=z_k} f(z)\neq 0\).

Since poles are isolated by definition, the set \(P_f\) of poles of \(f\) is a discrete subset of \({\mathbf{CP}}^1\), which is compact. Note \(P_f\) is closed because being holomorphic is an open condition. Any discrete closed subset of a compact space is discrete and compact, thus necessarily finite.

Applying this to any bounded \(\Omega \subseteq {\mathbf{C}}\), there can only be finitely many poles in any disc of radius \(R\). If there are infinitely many poles, this forces them to accumulate on \(z=0\infty\). A limit point of a sequence of poles of \(f\) is a limit point of a sequence of zeros of \(g\coloneqq 1/f\), making it an essential singularity for both.

Any such \(f\) can only have finitely many poles, so enumerate them as \(\left\{{z_k}\right\}_{k\leq n}\). Write \(P_k\) for the principal part of \(f\) at \(z_k\), so there is a decomposition \begin{align*} f(z) = \sum_{k \leq n} P_k(z) + Q(z) ,\end{align*} where \(Q(z)\) is now entire. Note that \(f(z)-Q(z)\) is evidently a rational function, and the claim is that \(Q\) is constant. Indeed, \({\mathbf{CP}}^1\) is compact and \(g\) is continuous, thus bounded, so Liouville applies. Thus \(f(z) = \sum_{k\leq n}P_k(z) + c\) is rational.

Heuristic: \(\sin(z)\) has a zero of order 1, so the \(z\) in the denominator exactly cancels it. Explicitly, this is evident from the Laurent expansion about zero: \begin{align*} z^{-1}\sin(z) = z^{-1}\qty{ z - {z^3 \over 3!} + {z^5\over 5!} - \cdots} = 1 - {z^2\over 3!} + {z^4 \over 5!} - \cdots ,\end{align*} which has no factors of \(z^{-k}\). So \(z=0\) is a removable singularity.

Note \(e^z = 1\) when \(z=z_k\coloneqq 2\pi k\) for \(k\in {\mathbf{Z}}\), and the claim is that these are all poles of order 1 of \(f(z)\). These are clearly poles of some order, since they are zeros of \(1/f\), and the order will be the smallest \(n\) for which \(\lim_{z\to z_k}(z-z_k)^n f(z)\) exists. Start by computing the first: \begin{align*} \lim_{z\to z_k}(z-z_k)f(z) = \lim_{z\to z_k} {z-z_k\over e^z - 1} \overset{\text{LH}}{=} \lim_{z\to z_k} {1\over e^z} = e^{-z_k} = 1 .\end{align*}

Write \(f(z) = \sum_{k\geq 0}c_k z^k\), so \(g(z) \coloneqq f(1/z) = \sum_{k\geq 0} c_k z^{-k}\). Since \(z=0\) is a pole of \(g\), \(c_k = 0\) for all \(k\geq m\) for \(m\) the order of the pole, so \(f(z) = \sum_{0\leq k\leq m}c_k z^k\) is a polynomial of degree at most \(m\).

Toward a contradiction, fix \(w\) and suppose no such sequence exists. Then for every sequence \(w_k\to z_0\), there is an \({\varepsilon}\) such that \({\left\lvert {f(w_k) - w} \right\rvert} \geq {\varepsilon}\) for all \(k\). So the function \(g(z) \coloneqq{1\over f(z) - w}\) has no poles in \(D_r(z_0)\setminus\left\{{ z_0 }\right\}\), and since each \(z_k\) is a pole of \(f\), each is a zero of \(g\). If \(z_0\) is a singularity, since \({\left\lvert {g(z)} \right\rvert} \leq {\varepsilon}\), it is removable and thus \(g\) can be extended holomorphically over \(z_0\). By continuity, since \(z_k\to z_0\) with \(g(z_k) = 0\), we have \(g(z_0) = 0\). By the identity principle, \(g\equiv 0\), which means that every \(z\in D_r(z_0)\setminus\left\{{ z_0 }\right\}\) is a zero of \(g\) and thus a pole of \(f\). But this contradicts that \(f\) is holomorphic on \(D_r(z_0)\setminus\left\{{ z_k }\right\}\). \(\contradiction\)

• For \(a\in {\mathbf{R}}_{< 0}\): take \(z_k\coloneqq{1\over \operatorname{Log}(a) + 2\pi i k - {\pi i \over 2}}\)
  Then \(f(z_k) = a\) for all \(k\) but \(z_k\to 0\).
• For \(a=0\): take \(z_k = -1/k\).
• For \(a=\infty\), take \(z_k = 1/k\).
• For anything else, take \(z_k \coloneqq{1\over \operatorname{Log}(a) + 2\pi i n}\) if \(a \in {\mathbf{R}}_{\geq 0}\). Again \(f(z_k) = a\) for all \(k\) but \(z_k\to 0\).

Note that writing a single function for each singularity and taking a product might work, except that there may be unforeseen cancellation of zeros of one with poles of another, or some might become removable. A surefire way is to take a sum, e.g.  \begin{align*} f(z) = e^{1\over z-1} + e^{1\over z} + {1\over z - (1-i)} + {1\over (z - (1+i) )^2 } + { z-7 \over \sin(z-7) } .\end{align*}

It can not be a pole: otherwise \({\left\lvert {f(z)} \right\rvert}\to \infty\) as \(z\to z_0\), but \(f(z) = 0\) infinitely often in every neighborhood of \(z_0\) since the \(z_k\) accumulate on it. It can not be removable: otherwise \(f\) extends holomorphically over \(z_0\), and continuity forces \(f(z_k) \to 0\) as \(z_k\to z_0\). But then \(f = 0\) on a set with an accumulation point, making \(f \equiv 0\) by the identity principle.

Note that the singularities are \begin{align*} z = 0, \pi, -\pi .\end{align*}

That \(z=0\) is removable: \begin{align*} \lim_{z\to 0} f(z) &= \lim_{z\to 0} {z-\sin(z) \over z\sin(z)} \\ &\overset{\text{LH}}{=} \lim_{z\to 0} {1 - \cos(z) \over \sin(z) + z\cos(z)} \\ &\overset{\text{LH}}{=} \lim_{z\to 0} {\sin(z) \over \cos(z) + \cos(z) -z\sin(z) } \\ &= 0 < \infty ,\end{align*} so in particular \(f\) is bounded in a neighborhood of \(z=0\), making it removable.

That \(z=\pi\) is removable: \begin{align*} \lim_{z\to \pi} f(z) &= \lim_{z\to \pi} {1\over \sin(z)} - {1\over z} + {1\over z-\pi} + {1\over z+\pi}\\ &= c_1 + \lim_{z\to \pi} {1\over \sin(z)} + {1\over z-\pi} \\ &= c_1 + \lim_{z\to \pi} { (z-\pi) -\sin(z) \over (z-\pi) \sin(z) }\\ &= c_1 + \lim_{w\to 0} { w -\sin(w + \pi) \over w \sin(w+\pi) } \qquad w\coloneqq z-\pi \\ &= c_1 - \lim_{w\to 0} { w + \sin(w) \over w \sin(w) } \\ &\overset{\text{LH}}{=} c_1 + 0 < \infty ,\end{align*} using the same L’Hopital argument as above. So this limit is bounded.

That \(z=-\pi\) is removable: \begin{align*} \lim_{z\to \pi} f(z) &= \lim_{z\to -\pi} {1\over \sin(z)} - {1\over z} + {1\over z-\pi} + {1\over z+\pi}\\ &= c_2 + \lim_{z\to -\pi} {1\over \sin(z)} + {1\over z+\pi}\\ &= c_2 + \lim_{z\to -\pi} {(z+\pi) - \sin(z) \over (z+\pi) \sin(z) } \\ &= c_2 - \lim_{z\to -\pi} {w + \sin(w) \over w \sin(w) } \qquad w \coloneqq z+\pi \\ &= c_2 + 0 < \infty ,\end{align*} again by the same argument.

For a Laurent expansion about \(z=0\), note \begin{align*} {1\over \sin(z) } &= {1\over z + c_3 z^3 + c_5 z^5 + { \mathsf{O}}(z^7)} \\ &= z^{-1}( 1 + c_3z^2 + (c_3^2-c_5)z^4 + { \mathsf{O}}(z^6)) \\ &= z^{-1}+ {1\over 3!} z + \qty{ \qty{1\over 3!}^2 - {1\over 5!} }z^3 + { \mathsf{O}}(z^5) \\ &= z^{-1}+ {1\over 6}z + {7\over 360}z^3 + { \mathsf{O}}(z^5) .\end{align*} and \begin{align*} {2z\over z^2-\pi^2} &= - {2z\over \pi^2} {1\over 1 - \qty{z\over \pi}^2 } \\ &= -{2z\over \pi^2}\sum_{k\geq 0}\qty{z\over \pi}^{2k} \\ &= -{2z\over \pi^2}\qty{1 + {1\over \pi^2} z^2 + {1\over \pi^4}z^4 + { \mathsf{O}}(z^6) } \\ &= -{2\over \pi^2}z -{2\over \pi^4}z^3 - {2\over \pi^6} z^5 - { \mathsf{O}}(z^7) ,\end{align*} so ` \begin{align*} f(z) &= \qty{z^{-1}+ {1\over 6}z + {7\over 360}z^3 + { \mathsf{O}}(z^5)} + \qty{-{2\over \pi^2}z -{2\over \pi^4}z^3 - {2\over \pi^6} z^5 - { \mathsf{O}}(z^7)}

• z^{-1}\ &= \qty{ {1\over 6} - {2\over \pi^2}}z + \qty{{7\over 360} - {2\over \pi^4} }z^3 + { \mathsf{O}}(z^5) .\end{align*} `{=html}

Write \begin{align*} f(z) = {z-\sin(z) \over z\sin(z)} - {2z\over z^2-\pi^2} ,\end{align*}

For \(z=0\), the 2nd term doesn’t contribute to zero/pole order. For the first, take an expansion: \begin{align*} f_1(z) &= {z - \qty{ z + c_3z^3 + { \mathsf{O}}(z^5)} \over z \qty{z + c_3z^3 + { \mathsf{O}}(z^5)} } \\ \\ &= { -c_3z^3 + { \mathsf{O}}(z^5)\over z^2 + { \mathsf{O}}(z^4) } ,\end{align*} so there is a zero of order 3 in the numerator and of order 2 in the denominator, making the singularity removable. A similar argument works at \(z=\pm \pi\).

It suffices to show that \(f'\) is bounded in a neighborhood of \(z_0\). Since \(z_0\) is a removable singularity of \(f\), there is a neighborhood \({\mathbb{D}}_R(a)\) on which \({\left\lvert {f(z)} \right\rvert} \leq M\) is bounded. Using the Cauchy estimates, \begin{align*} {\left\lvert {f'(z_0)} \right\rvert} &\leq {1\over 2\pi } \oint_{{\left\lvert {z-a} \right\rvert} = R } {{\left\lvert {f(z)} \right\rvert} \over {\left\lvert {z-z_0} \right\rvert}^2 } \,dz\\ &\leq {1\over 2\pi } \oint_{{\left\lvert {z-a} \right\rvert} = R } MR^{-2} \,dz\\ &= {1\over 2\pi } MR^{-2} \cdot 2\pi R \\ &= MR^{-1}< \infty .\end{align*}

For the converse, if \(z_0\) is removable for \(f'\), write \(F'\) for the holomorphic extension of \(f'\) over \({\mathbb{D}}_{\varepsilon}(a)\) which exists by Riemann’s removable singularity theorem. Since \(F'\) is holomorphic, it has a primitive \(F(z) \coloneqq\int_{w}^z F'(\xi) \,d\xi\) for any point \(w\) in this region. Now \(G\coloneqq F' - f' \equiv 0\) on \({\mathbb{D}}_{\varepsilon}^*(a)\) making \(G\equiv c\) constant, so \(f(z) = F(z) + c\). In particular, \begin{align*} \lim_{z\to a} f(z) = \lim_{z\to a} F(z) + c = \lim_{z\to a} \int_w^z F' \,dz+ c = F'(a) + c < \infty ,\end{align*} so \(a\) is removable for \(f\).

Define \(F(z) \coloneqq z^k f^{(k)}(z)\) and note that \({\left\lvert {F(z)} \right\rvert} \leq M\) on \({\mathbb{D}}\setminus\left\{{0}\right\}\).

If \(f\) has an essential singularity at \(z=0\), then so does \(F\) by considering power series expansions: \begin{align*} f(z) = \sum_{k\in {\mathbf{Z}}} c_k z^k \implies z^m f^{(m)}(z) = \sum_{k\leq 1} \tilde c_k z^{-k} + \sum_{k\geq m}\tilde c_{k}z^{k} ,\end{align*} which will still have infinitely many terms in its principal part at \(0\). However, if \(F\) had an essential singularity, the image of \(F\) in a neighborhood of \(0\) would be dense in \({\mathbf{C}}\) by Casorati-Weierstrass, contradicting that its image is bounded (by \(M\)).

Suppose instead \(z=0\) is a pole of order \(\ell\) of \(f\), so \({\left\lvert {f(z)} \right\rvert}\to \infty\) as \(z\to 0\). Then again by considering power series expansions, \(z=0\) remains a pole of \(F\), now of order at worst \(\ell\): \begin{align*} f(z) = { \mathsf{O}}(z^{\ell}) \implies z^m f^{(m)}(z) \approx z^m \cdot { \mathsf{O}}(z^{\ell - m}) = { \mathsf{O}}(z^\ell) .\end{align*} But if this is an order \(\ell\) pole of \(F\), then \(\lim_{z\to 0} {\left\lvert {F(z)} \right\rvert} = \infty\) and \(\lim_{z\to 0} z^\ell F(z))\) is finite and nonzero. Apply the assumed bound yields the last contradiction: \begin{align*} z^{\ell}F(z) = z^{\ell + m}f(z) \leq z^{\ell + m} \cdot Mz^{-m} = z^{\ell} \overset{z\to 0}\longrightarrow 0 .\end{align*} \(\contradiction\)

Using the inequality, \begin{align*} {\left\lvert {(z-0)f(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}^{1\over 2}\overset{{\left\lvert {z} \right\rvert}\to 0}\longrightarrow 0 ,\end{align*} so \(z=0\) is removable by Riemann’s removable singularity theorem.

Suppose \(f\) is not constant. If \(z=\infty\) is removable, \(f\) is bounded in a neighborhood of \(\infty\), say by \(M_1\) on \({\left\lvert {z} \right\rvert} > R\). Now \({\left\lvert {z} \right\rvert} \leq R\) is a closed and bounded set, thus compact, and since \(f\) is continuous here it is bounded by the extreme value theorem, say by \(M_2\). Then \({\left\lvert {f(z)} \right\rvert} \leq \max(M_1, M_2)\) on \({\mathbf{C}}\) is entire and bounded, thus constant by Liouville, a contradiction. \(\contradiction\)

Conversely, if \(f\) is constant, \(f\) is trivially bounded in every neighborhood of \(\infty\), making it a removable singularity.

Since \(f\) is entire, \(f(z) = \sum_{k\geq 0 } c_k z^k\). Expanding about \(z_0=\infty\), we have \(f(1/z) = \sum_{k\geq 0} c_k z^{-k} = c_0 + {c_1\over z} + \cdots\). If \(z_0=\infty\) is a pole of order \(m\), then \(c_m\neq 0\) but \(c_{>m} = 0\), which forces \(f(z) = \sum_{0\leq k \leq m} c_k z^k\) to be a polynomial of degree \(m\).