# Singularities

## Spring 2020 HW 3.3 #complex/exercise/completed

Let $$P, Q$$ be polynomials with no common zeros. Assume $$a$$ is a root of $$Q$$. Find the principal part of $$P/Q$$ at $$z=a$$ in terms of $$P$$ and $$Q$$ if $$a$$ is

• a simple root, and
• a double root.

Write \begin{align*} P(z) &= \prod_{k\leq n} (z-a_k) \\ Q(z) &= \prod_{k\leq m}(z-b_k) \\ Q_j(z) &= \prod_{k\neq j}(z-b_k) = {Q(z) \over z-z_j} .\end{align*}

For $$b_\ell$$ a simple pole, \begin{align*} {P(z) \over Q(z) } = {1\over z-b_\ell} {P(z) \over Q_\ell(z)} &= {1\over z-b_\ell}\qty{c_0 + c_1(z-b_\ell) + c_2(z-b_\ell)^2 + \cdots} \\ &= {c_0 \over z-b_\ell} + c_1 + { \mathsf{O}} (z-b_\ell) \\ & \coloneqq{P_{b_\ell}(z)} + c_1 + { \mathsf{O}} (z-b_\ell) ,\end{align*} so the principal part at $$z=z_\ell$$ is given by \begin{align*} P_{z_\ell}(z) = {c_0 \over z-b_\ell} = {P(z) \over Q_\ell(z)}\Big|_{z=b_\ell} = \lim_{z\to z_\ell} {(z-b_\ell) P(z) \over Q(z)} .\end{align*}

For $$b_\ell$$ a double pole, \begin{align*} {P(z) \over Q(z) } &= {1\over (z-b_\ell)^2 } {(z-z_\ell)^2P(z) \over Q(z) } \\ &= {1\over (z-b_\ell)^2}\qty{ d_0 + d_1(z-b_\ell) + d_2(z-b_\ell)^2 } \\ &= {d_0 \over (z-b_\ell)^2} + {d_1\over z-z_\ell} + d_2 + { \mathsf{O}} (z-b_\ell) \\ &\coloneqq P_{b_\ell}(z) + d_2 + { \mathsf{O}} (z-b_\ell) .\end{align*} To extract the $$d_1$$ coefficient, note that \begin{align*} {(z-b_\ell)^2 P(z) \over Q(z)} &= d_0 + d_1(z-b_\ell) + \cdots \\ \implies {\frac{\partial }{\partial z}\,} {(z-b_\ell)^2 P(z) \over Q(z)} &= d_1 + 2d_2(z-b_\ell) + \cdots ,\end{align*} so \begin{align*} d_0 &= \lim_{z\to b_\ell} { (z-b_\ell)^2 P(z) \over Q(z) } \\ d_1 &= \lim_{z\to b_\ell} {\frac{\partial }{\partial z}\,} {(z-b_\ell)^2 P(z) \over Q(z) } \\ P_{b_\ell} &= {d_0 \over (z-b_\ell)^2} + {d_1\over z-b_\ell} .\end{align*}

## Spring 2020.4 #complex/qual/completed

Suppose that $$f$$ is holomorphic in an open set containing the closed unit disc, except for a simple pole at $$z=1$$. Let $$f(z)=\sum_{n=1}^{\infty} c_{n} z^{n}$$ denote the power series in the open unit disc. Show that \begin{align*} \lim _{n \rightarrow \infty} c_{n}=-\lim _{z \rightarrow 1}(z-1) f(z) .\end{align*}

Compute the series expansion of the RHS: \begin{align*} (z-1) f(z) &= (z-1) \sum_{n\geq 1} c_n z^k \\ &= -c_1z + \sum_{n\geq 2} (c_{n-1} - c_n) z^n \\ &\overset{z\to 1}\longrightarrow -c_1 + \sum_{n\geq 2} c_{n-1} - c_n \\ &\coloneqq\lim_{N\to\infty} -c_1 z + \sum_{n=2}^N c_{n-1} - c_n \\ &= \lim_{N\to\infty} -c_N ,\end{align*} where we’ve used that the sum is telescoping.

## Entire functions with poles at $$\infty$$#complex/exercise/completed

Find all entire functions with have poles at $$\infty$$.

If $$f$$ is entire, write $$f(z) = \sum_{k\geq 0}c_k z^k$$ and $$g(z) \coloneqq f(1/z) = \sum_{k\geq 0}c_k z^{-k}$$. If $$z=\infty$$ is a pole of order $$m$$ of $$f$$, $$z=0$$ is a pole of order $$m$$ of $$g$$, so \begin{align*} g(z) = \sum_{0\leq k \leq m}c_k z^{-k} \implies f(z) = \sum_{0\leq k \leq m}c_k z^k ,\end{align*} making $$f$$ a polynomial of degree at most $$m$$.

## Functions with specified poles (including at $$\infty$$) #complex/exercise/completed

Find all functions on the Riemann sphere that have a simple pole at $$z=2$$ and a double pole at $$z=\infty$$, but are analytic elsewhere.

Write $$f(z) = P_2(z) + g(z)$$ where $$P_2$$ is the principal part of $$f$$ at $$z=2$$ and $$g$$ is holomorphic at $$z=2$$. Then $$g$$ is an entire function with a double pole at $$\infty$$, and is thus a polynomial of degree at most $$2$$, so $$g(z) = c_2z^2 + c_1 z + c_0$$. Since the pole of $$f$$ at $$z=2$$ is simple, $$P_2(z) = \sum_{k\geq -1} d_k (z-2)^k$$. Combining these, we can write \begin{align*} f(z) = d_{-1}(z-2)^{-1}+ \sum_{0\leq k\leq 3} (d_k + c_k)(z-2)^k + \sum_{k\geq 3}d_k (z-2)^k .\end{align*} However, if $$d_k\neq 0$$ for any $$k\geq 3$$, this results in a higher order pole at $$\infty$$, so $$f$$ must be of the form \begin{align*} f(z) = d_{-1}(z-2)^{-1}+ \sum_{0\leq k\leq 3} (d_k + c_k)(z-2)^k .\end{align*}

## Entire functions with singularities at $$\infty$$#complex/exercise/completed

Let $$f$$ be entire, and discuss (with proofs and examples) the types of singularities $$f$$ might have (removable, pole, or essential) at $$z=\infty$$ in the following cases:

• $$f$$ has at most finitely many zeros in $${\mathbb{C}}$$.
• $$f$$ has infinitely many zeros in $${\mathbb{C}}$$.

Write $$f(z) = \sum_{k\geq 0} c_k z^k$$ since it is entire.

• If $$f$$ has finitely many zeros, $$f$$ is nonconstant and entire, and thus unbounded by Liouville. If $$f$$ is nonconstant, $$z=\infty$$ can not be removable, since this would force $$f$$ to be constant. So $$z=\infty$$ can be a pole or an essential singularity. Both possibilities can occur: if $$f$$ is a polynomial, it is entire with finitely many zeros and a pole at $$z=\infty$$. Taking $$f(z)= e^z$$ has no zeros and an essential singularity at $$z=\infty$$.

• If $$f$$ has infinitely many zeros, if $$f$$ is nonconstant then infinitely many $$c_k$$ are nonzero – otherwise $$f$$ is a polynomial and can only have finitely many zeros. Then $$g(z) \coloneqq f(1/z) = \sum_{k\geq 0}{c_k\over z^k}$$ has infinitely many nonzero terms, making $$z=0$$ an essential singularity for $$g$$ and $$z=\infty$$ essential for $$f$$.

## Sum formula for $$\sin^2$$#complex/exercise/completed

Define \begin{align*} f(z) &= {\pi^2 \over \sin^2 \qty{\pi z} } \\ g(z) &= \sum_{n\in {\mathbb{Z}}} {1\over (z-n)^2} .\end{align*}

• Show that $$f$$ and $$g$$ have the same singularities in $${\mathbb{C}}$$.
• Show that $$f$$ and $$g$$ have the same singular parts at each of their singularities.
• Show that $$f, g$$ each have period one and approach zero uniformly on $$0\leq x \leq 1$$ as $${\left\lvert {y} \right\rvert}\to \infty$$.
• Conclude that $$f = g$$.

Part 1: This is clear: $$\sin^2(\pi z) = 0 \iff z = k$$ for $$k\in {\mathbb{Z}}$$, and this is a pole of order 2 for $$f$$. Every $$k\in {\mathbb{Z}}$$ is visibly an order 2 pole of $$g$$.

Part 2: By periodicity, it suffices to consider the singularity at $$z_0 = 0$$. Expanding $$\sin(\pi z) = \pi z - {1\over 3!}(\pi z)^3 + {1\over 5!} (\pi z)^5 + \cdots$$ and considering $$\sin(\pi z)^2$$ shows that $$z=0$$ is a pole of order 2. So $$z^2f(z)$$ has a removable singularity at $$z=0$$, and can be expanded: \begin{align*} z^2f(z) &= \qty{\pi z\over \sin(\pi z)}^2 \\ &= (\pi z)^2 \qty{ (\pi z) ^{-1}+ {1\over 3!}(\pi z) + {7\over 360} (\pi z^3) + \cdots}^2 \\ &= (\pi z)^2 \qty{ (\pi z)^{-2} + { \mathsf{O}} (1) } \\ &= 1 + { \mathsf{O}} (z^2) \\ \implies f(z) &= z^{-2} + { \mathsf{O}} (1) ,\end{align*} so the singular part of $$f$$ at $$z=0$$ is $$z^{-2}$$. This coincides with the $${1\over z^2}$$ term in $$g$$. The remaining principal parts at $$z=k$$ are $${1\over (z-k)^2},$$ using the fact that $$f(z+1) = f(z)$$, so $$f(k) = f(0)$$ and the Laurent expansions are gotten by substituting $$z-k$$ in for $$z$$ everywhere.

Part 3: Periodicity is clear for $$f$$. For $$g$$, \begin{align*} g(z+1) = \sum_{k\in {\mathbb{Z}}} ((z-1)-k)^{-2} = \sum_{k'\in {\mathbb{Z}}} (z-k)^{-2} ,\end{align*} where $$k' \coloneqq k+1$$, and the equality is true since both sums run over all of $${\mathbb{Z}}$$.

For convergence: take $$z=it$$, then for $$f$$ \begin{align*} f(it) \sim \csc^2(i\pi t) &\sim \qty{ e^{i\pi (it) } - e^{-i\pi (it)}}^{-2} \\ &= \qty{e^{-\pi t} - e^{\pi t}}^{-2} \\ &\leq {1\over e^{-\pi t} + e^{\pi t} } \\ &\sim e^{-\pi t} \\ &\to 0 ,\end{align*} using the reverse triangle inequality and that the $$e^{-\pi t}$$ term in the denominator is negligible for large $$t$$.

For $$g$$, \begin{align*} g(it) &\sim t^{-2} + \sum_{k\geq 1} (t^2 + k^2)^{-1}\\ &\leq t^{-2} + \sum_{1\leq k \leq N}(t^2 + k^2)^{-1}+ \sum_{k\geq N}(t^2 + k^2)^{-1}\\ &\leq t^{-2} + \sum_{1\leq k \leq N}(t\cdot k^2)^{-1}+ \sum_{k\geq N}(k^2)^{-1}\\ &\leq t^{-2} + t^{-1}\sum_{1\leq k \leq N}(k^2)^{-1}+ \sum_{k\geq N}(k^2)^{-1}\\ &\overset{N\to\infty\implies t\to\infty}\longrightarrow 0 ,\end{align*} where given $$N$$ we can pick $$t$$ large enough so that $$t^2 + k^2 \geq tk^2$$ for all $$k\leq N$$. These converge to zero as $$N\to\infty$$ since $$\sum k^{-2} < \infty$$, making the last term the tail of a convergent sum.

Part 4: Since $$f,g$$ uniformly converge to zero on the strip $$0<\Re(x) < 1$$, they are bounded on this strip. Since this is a fundamental domain for their periods, they are bounded on $${\mathbb{C}}$$. Write $$h\coloneqq f-g$$, then $$h$$ is entire since $$f,g$$ have the same singular parts, and bounded since $${\left\lvert {h} \right\rvert}\leq {\left\lvert {f} \right\rvert} + {\left\lvert {g} \right\rvert}$$. By Liouville, $$h$$ is constant with $$\lim_{t\to\infty} h(it) = 0$$, so $$h\equiv 0$$ and $$f\equiv g$$.

## Spring 2020 HW 3.4, Tie’s Extra Questions: Fall 2015 #complex/qual/completed

Let $$f(z)$$ be a non-constant analytic function in $$|z|>0$$ such that $$f(z_n) = 0$$ for infinite many points $$z_n$$ with $$\lim_{n \rightarrow \infty} z_n =0$$.

Show that $$z=0$$ is an essential singularity for $$f(z)$$.

Hint: an example of such a function is $$f(z) = \sin (1/z)$$.

Note that $$z=0$$ can not be a removable singularity, since then $$f$$ would extend to a holomorphic function over $$z=0$$, and by continuity $$0 = \lim f(z_n) = f(\lim z_n) = f(0)$$. By the identity principle, this would force $$f\equiv 0$$, contradicting that $$f$$ is nonconstant.

It can not be a pole, because then $$f(z_n)\to \infty$$, but $${\left\lvert {f(z_n)} \right\rvert} = 0 < {\varepsilon}$$ for any $${\varepsilon}$$ infinitely many times.