# Exercises: Singularities

## General

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*}

## Zeros

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

## Poles

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 $$m$$.

#complex/exercise/completed