Exercises: Singularities


Prove that the following statements or true, or find a counterexample:

  • If \(f,g\) have a pole at \(a\), then \(f+g\) has a pole at \(a\).
  • If \(f,g\) have a pole at \(a\), then \(fg\) has a pole at \(a\).
  • If \(f\) has an essential singularity at \(z_0\) at \(g\) is has a pole at \(z_0\), then \(z_0\) is an essential singularity for \(f+g\).
  • If \(f\) has a pole of order \(N\) at \(z_0\) then \(f^2\) has a pole of order \(2N\) at \(z_0\).


  • 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.

Classify the singularities of \begin{align*} f(z) = {z^3+1\over z^2(z+1)} .\end{align*}


Showing a pole \(z_0\) of \(f\) is order \(n\): show that \(z_0\) is a zero of order \(n\) of \(1/f\), i.e. \(1/f = (z-z_0)^nh(z)\) with \(h\) nonvanishing in a neighborhood of \(z_0\).

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 \(z^{-1}\).

Classify the singularities at \(z=0\) of the following \begin{align*} f_1(z) &= {\operatorname{Log}(1+z) \sin(z) \over z^2} \\ f_2(z) &= e^{\sin\qty{1\over z}} \\ f_3(z) &= {1+z \over e^z-1} .\end{align*}


\(f_1\): 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*}


If \(f\) is holomorphic on \(\Omega\) and not identically zero, then \(f^{-1}(0) \cap\Omega\) is discrete.


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.

Show that the complex zeros of \(f(z) \coloneqq\sin(\pi z)\) are exactly \({\mathbf{Z}}\), and each is order 1. Calculate the residue of \(1/\sin(\pi x)\) at \(z=n\in {\mathbf{Z}}\).


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*}

Orders of poles/zeros

Find the orders of zeros of the following functions:

  • \((e^z-1)^3\)


  • \(z=0\) 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.

Determine the order of the pole of

  • \({1\over z\sin(z)}\) at \(z_0 = 0\).
  • \({e^{z^2}-1\over z^4}\) at \(z_0=0\)


  • 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*}

Show that if \(z_0\) is a pole of order \(n\) of \(f\), then it is a pole of order \(n+k\) for \(f^{(k)}\).


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 \(N+1\).

Let \(f\) be an elliptic function and \(P\) be its fundamental parallelogram. Supposing that \(f\) is nonconstant, show that \(f\) has at least two poles in \(P\) (counted with multiplicity).


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. \(\sum_{k} \mathop{\mathrm{Res}}_{z=z_k} f(z)\neq 0\).


Show that a meromorphic function on \({\mathbf{CP}}^1\) can have only finitely many poles. Show that moreover if \(f\) is meromorphic on \({\mathbf{C}}\) with infinitely many poles, then the poles must accumulate on an essential singularity at \(z=\infty\).


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.

Show that the only meromorphic functions on \({\mathbf{CP}}^1\) are rational functions.


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.

Show that \(\sin(z)/z\) has no poles.


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.

Classify the singularities and compute the residues at any poles of the following function: \begin{align*} f(z) \coloneqq{1\over e^z - 1} .\end{align*}


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*}

Show that if \(f\) is entire and \(f(1/z)\) has a pole at \(z=0\), then \(f\) is a polynomial.


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 \(m\).

Essential Singularities

Let \(f\) be holomorphic in \(0 < {\left\lvert {z-z_0} \right\rvert} < r\), minus a sequence of poles \(\left\{{z_k}\right\} \to z_0\). Show that for any \(w\in {\mathbf{C}}\), there is a sequence \(\left\{{w_k}\right\}\to z_0\) with \(f(w_k)\to w\).


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 \(D_r(z_0)\setminus\left\{{ z_k }\right\}\). \(\contradiction\)

Fix \(a\in {\mathbf{C}}\cup\left\{{\infty}\right\}\) and let \(f(z) \coloneqq e^{1\over z^2}\). Find a sequence \(z_k\to 0\) such that \(f(z_k) \overset{k\to\infty}\longrightarrow a\)


  • 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 \(z_k\to 0\).

Determine a function with

  • An essential singularity at \(z=1\)
  • An essential singularity at \(z=0\)
  • A pole of order 1 at \(z=1-i\)
  • A pole of order 2 at \(z=1+i\)
  • A removable singularity at \(z=7\)


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*}

Suppose \(f\not\equiv 0\) is holomorphic on \(\Omega\setminus\left\{{ z_0 }\right\}\) with a sequence of zeros \(z_k\) limiting to \(z_0\). Show that \(z_0\) is an essential singularity of \(f\).


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.

Removable Singularities

Consider \begin{align*} f(z) \coloneqq{1\over \sin(z)} - {1\over z} + {2z\over z^2-\pi^2} .\end{align*} Show that on \({\left\lvert {z} \right\rvert} < 2\pi\), all singularities are removable, and find a Laurent expansion about \(z=0\).


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 \(z=\pm \pi\).

Suppose \(f\) is meromorphic. Show that if \(z_0\) is a removable singularity of \(f\), then it is also a removable singularity of \(f'\). Conversely, if \(z_0\) is removable for \(f'\), then it is also removable for \(f\).


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 \(f\).

Suppose \(f\) is holomorphic on \({\mathbb{D}}\setminus\left\{{0}\right\}\) and there exist \(M, k\) such that \begin{align*} {\left\lvert {f^{(k)}(z)} \right\rvert} \leq {M\over {\left\lvert {z} \right\rvert}^k} && \forall 0 < {\left\lvert {z} \right\rvert} < 1 .\end{align*}

Show that if \(f\) has a singularity at \(z=0\), then it must be removable.


  • \({\frac{\partial }{\partial z}\,}\) is a left-shift on power series, \(z^m\) is a right-shift.
  • \(f'\) has the same poles as \(f\), possibly with worse order due to the left-shift.
    • In general, if \(z_0\) is an order \(\ell\) pole of \(f\), then it is at least an order \(\ell + m\) pole of \(f^{(m)}\).

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*} \(\contradiction\)

Suppose \(f\) is holomorphic with \(z_0 = 0\) an isolated singularity, and suppose there is some neighborhood of \(0\) on which \begin{align*} {\left\lvert {f(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}^{-{ 1\over 2}} .\end{align*} Show that \(z_0\) is removable.

Warning: Riemann’s removable singularity theorem won’t apply to \(z^{1\over 2}f(z)\) since \(z^{1\over 2}\) is highly singular at \(z=0\).


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.

Singularities at Infinity

Let \(f\) be entire. Show that \(f\) has a removable singularity at \(z_0 = \infty\) iff \(f\) is constant.


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.

Characterize all entire functions with a pole of order \(m\) at \(\infty\).


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 \(m\).