Author Topic: TT2 Problem 3  (Read 4049 times)

Victor Ivrii

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TT2 Problem 3
« on: November 24, 2018, 04:29:17 AM »
Find all singular points of
$$f(z)=z^2(z^2-\pi^2)\cot^2(z)$$
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 $g(w)=f(1/w)$?).

Jihang Yu

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Re: TT2 Problem 3
« Reply #1 on: November 24, 2018, 04:48:57 AM »
We have

f(z) = z^2(z^2-\pi^2)\cot^2(z) \\
f(z) = z^2(z^2-\pi^2)\frac{\cos^2(z)}{sin^2(z)} \Rightarrow \sin^2(z) \neq 0 \Rightarrow \sin(z) \neq 0 \Rightarrow z \neq n\pi, n \in \mathbb{Z}

So the singular points are $z_0 = n\pi, n \in \mathbb{Z}$.

Now consider the part

\frac{\cos^2(z)}{\sin^2(z)} = \frac{1-\sin^2(z)}{\sin^2(z)} =\frac{1}{\sin^2(z)} - 1\\
\lim_{z \to z_0} \frac{\cos^2(z)}{\sin^2(z)} = \infty - 1 = \infty

Notice when $z_0 = n\pi, n=0$, we have $z^2=0$.
When $z_0 = n\pi, n=\pm1$, we have $(z^2 -\pi^2) = 0.$ Therefore for $z_0=n\pi, n=0,\pm1$

\lim_{z \to z_0} f(z) = 0

So there are three removable singularities.

For $z_0=n\pi, n \neq 0,\pm1$

\lim_{z \to z_0} f(z) = \infty

So there are all poles of order 2.
« Last Edit: November 24, 2018, 06:27:09 AM by Jihang Yu »

Jihang Yu

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Re: TT2 Problem 3
« Reply #2 on: November 24, 2018, 05:21:40 AM »
Now let $z=\frac{1}{w}$, we have

f(z)=f(\frac{1}{w})=\frac{1}{w^2}(\frac{1}{w^2}-\pi^2)\frac{\cos^2(\frac{1}{w})}{\sin^2(\frac{1}{w})} \\
\lim_{w \to 0} f(\frac{1}{w}) = \infty \cdot \infty \cdot a \qquad a = \infty, a = constant, a = 0 \\
\lim_{w \to 0} f(\frac{1}{w}) = \infty,0 \Rightarrow \lim_{z \to \infty} f(z) = \infty,0

So it is an not isolated singularity
« Last Edit: November 24, 2018, 06:28:05 AM by Jihang Yu »

Ende Jin

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Re: TT2 Problem 3
« Reply #3 on: November 25, 2018, 05:30:24 PM »
(Note: At $z = \pi$ or $z = -\pi$, $f(z)$ has pole of order 1, since
$\lim_{z \rightarrow \pi} f(z) = (l'hopital) \lim_{z \rightarrow \pi} \frac{(4z^3-2z\pi^2) \cos^2(z) - \sin(2z)(z^4 - z^2\pi^2)}{sin(2z)} = \infty$, which must be a pole
)

Complete Solution:
Since
$$f(z) = \frac{z^2(z+\pi)(z-\pi) \cos^2(z)}{\sin^2(z)}$$
thus,
$f(z)$ has singularities at $z = n\pi$ where $n \in \mathbb{Z}$.
and at each $n\pi$, $\sin^2(z)$ has zero of order 2.

Consider numerator, $z \mapsto z^2(z+\pi)(z-\pi)\cos^2(z)$ has zero of order 2 at $0\pi$, zero of order 1 at $\pi$ and $-\pi$, thus $f(z)$ has a removable singularities at 0, pole of order 1 at $\pi$ and $-\pi$, and pole of order two at $n\pi$ where $|n| \ge 2$.

Consider $g(z) := f(\frac{1}{z})$, since any disc centered at 0 in $g$ will include more than 1 singularities, that means, it is no the case that there exists a small ball, 0 is the only singularity in it. We can conclude $\infty$ is non-isolated singularity for $f$.
« Last Edit: November 26, 2018, 12:29:38 PM by Ende Jin »

Victor Ivrii

Ende is right with corrections at $z=\pm \pi$ and with $\infty$ to be non-isolated.