### Author Topic: TT2A Problem 3  (Read 2315 times)

#### Victor Ivrii

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##### TT2A Problem 3
« on: November 24, 2018, 04:55:52 AM »
Find all singular points of
$$f(z)=(z^2-1)\cot(\pi z^2)$$
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 $g(w)=f(1/w)$?).

#### Junya Zhang

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##### Re: TT2A Problem 3
« Reply #1 on: November 24, 2018, 08:55:19 AM »
$$f(z)=(z^2 - 1) \cot(\pi z^2)=(z+1)(z-1)\frac{\cos \pi z^2}{\sin \pi z^2}$$

$f$ has a singularity at $z$ iff $\sin \pi z^2 = 0$, this happens when $z^2 = k$ for $k \in \mathbb{Z}$.
Note that $\cos \pi z^2 = 0$ iff $z^2 = \frac{1}{2} + \mathbb{Z}$, which is mutually exclusive from zeros of $\sin \pi z^2 = 0$.

Case 1: $z = -1$
Since $(z+1)$ has a zero of order 1, $\sin \pi z^2 = 0$ has a zero of order 1, and no other factors have zeros at $z=-1$, then $f$ has a removable singularity at $z=-1$.

Case 2: $z = 0$
Since $\sin \pi z^2 = 0$ has a zero of order 2, and no other factors have zeros at $z=0$, then $f$ has a pole of order 2 at $z=0$ .

Case 3: $z = 1$
Since $(z-1)$ has a zero of order 1, $\sin \pi z^2 = 0$ has a zero of order 1, and no other factors have zeros at $z=1$, then $f$ has a removable singularity at $z=1$.

Case 4: $z = \pm\sqrt{k}$ for $k = -1, \pm2, \pm3,\cdots$
Since $\sin \pi z^2 = 0$ has a zero of order 1, and no other factors have zeros at these values, then $f$ has a simple pole at $z = \pm\sqrt{k}$ for $k = -1, \pm2, \pm3,\cdots$.

Case 5: singularity at $\infty$
Let $$g(w) = f(1/w) = \left(\frac{1}{w}+1\right)\left(\frac{1}{w}-1\right)\frac{\cos \left( \frac{\pi}{w^2}\right)}{\sin \left(\frac{\pi}{w^2}\right)}$$

Note that $g$ has singularities at $w=0$ and $w = \pm \sqrt\frac{1}{k}$ for $k \in \mathbb{Z}$. This implies that $\forall R>0, \exists w\in\mathbb{C}, 0<|w|<R$ such that $g$ is has a singularity at $w$, and thus $g$ is not analytic in $0<|w|<R$ for any $R>0$.
Therefore, the singularity of $g$ at $z=0$ is a non-isolated singularity. And thus, the singularity of $f$ at $\infty$ is a non-isolated singularity.
« Last Edit: November 24, 2018, 09:20:48 AM by Junya Zhang »