Author Topic: TT2B Problem 3  (Read 3393 times)

Victor Ivrii

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TT2B Problem 3
« on: November 24, 2018, 05:21:14 AM »
Find all singular points of
$$f(z)=\frac{\sin (\pi z)}{\sin(\pi z^3)}$$
and determine their types (removable, pole (in which case what is it's order), essential singularity, not isolated singularity, branching point).

In particular, determine singularity at $\infty$ (what kind of singularity we get at $w=0$ for $g(w)=f(1/w)$?).

Zixuan Miao

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Re: TT2B Problem 3
« Reply #1 on: November 24, 2018, 09:01:33 AM »
see attached.

Huanglei Ln

• Jr. Member
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Re: TT2B Problem 3
« Reply #2 on: November 25, 2018, 02:45:19 AM »
\begin{align*}
f(z)&=\frac{sin(\pi z)}{\pi z^3}\\
sin(\pi z^3)&=0\\
\Rightarrow \pi z^3&=k\pi,k\in Z\\
\Rightarrow z^3=k&z_0=\sqrt[3]{k}
\left\{
\begin{array}{lr}
k\ is\ a\ perfect\ cubic,\sqrt[3]{k}\in {z-\{0\}}&  \\
k\ is\ not\ a\ perfect\ cube& \\
k=0, &
\end{array}
\right.
\end{align*}

case1:

\begin{align*}
f(z)&=\underbrace{sin(\pi z)}_{g(z)}\underbrace{\frac{1}{\pi z^3}}_{h(z)}\\
\sqrt[3]{k}&=Z\Rightarrow z\in Z\Rightarrow g(\sqrt[3]{k})=0\\
g'(z)&=\pi cos(\pi z)\Rightarrow g'(\sqrt[3]{k})=\pm \pi\neq0
\end{align*}

$\Rightarrow$ g(z) has a zero of order 1 at $\sqrt[3]{k}$

\begin{align*}
\frac{1}{h(z)}&=sin(\pi z^3)\\
\frac{d}{dz}sin(\pi z^3)&=cos(\pi z^3)3\pi z^2
\end{align*}

$cos(\pi k)3\pi k^{\frac{2}{3}}\neq 0$

$\Rightarrow$ h(z) has a pole of order 1 at $\sqrt[3]{k}$

Thus, f(z)has a removable singularity at $z=\sqrt[3]{k}\in Z-\{0\}$.

case2:

\begin{align*}
\sqrt[3]{k}\notin Z \Rightarrow g(z_0)\neq0
\end{align*}

h(z) has a pole of order 1 at $\sqrt[3]{k}$ as shown in case1

Thus, f(z) has a pole of order 1 at$z=\sqrt[3]{k} \notin Z$.

case3:

$g(0)=0,g'(0)\neq 0,\Rightarrow g(z)$ has zero od order 1 at z=0

$\frac{d}{dz}sin(\pi z^3)=xos(\pi z^3)3\pi z^2$

$\frac{d^2}{dz^2}sin(\pi z^3)=cos(\pi z^3)6\pi z+3\pi z^2(-3z^2\pi sin(\pi z^3))$

$\frac{d^3}{dz^3}sin(\pi z^3)=cos(\pi z^3)6\pi +6\pi z(-3\pi z^2sin(\pi z^3)+\cdots$

Note $\frac{d^3}{sz^3}|_{z=0}\neq0\Rightarrow h(z)$ has pole of order 3 at z=0

Thus f(z) has pole of order 2 at zero.

Victor Ivrii

The singularities are obviously at $z=\sqrt[3]{n}$ with $n\in \mathbb{Z}$ and if $n\ne 0$ denominators have simple zeroes and numerator is not zero unless $n$ is a perfect cube. On the other hand, at $z=0$ numerator has a simple zero and denominator has a triple zero.
z_n=\sqrt[3]{n} \quad\text{is } \left\{\begin{aligned} &\text{removable singularity} &&\text{if n\in \mathbb{Z}, n\ne 0 is a perfect cube},\\ &\text{simple pole} &&\text{if n\in\mathbb{Z}, n\ne 0 is not a perfect cube},\\ &\text{double pole } &&\text{if }\ n=0. \end{aligned}\right.