Author Topic: Term Test 2 sample P3  (Read 1510 times)

Victor Ivrii

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Term Test 2 sample P3
« on: October 30, 2018, 05:25:41 AM »
Find all singular points of
$$
z^3 \tan  ( \pi z ) \cot^2  (\pi z ^2) $$
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 $g(w)=f(1/w)$?).

oighea

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Re: Term Test 2 sample P3
« Reply #1 on: November 04, 2018, 05:43:36 AM »
As $z^3 \tan(\pi z)\cot^2(\pi z^2)$ involves quotients of trigonometric functions, we obtain:

$\displaystyle f(z) = z^3 \frac{\sin(\pi z)}{\cos(\pi z)}\frac{\cos^2(\pi z^2)}{\sin^2(\pi z^2)}$.

Requirements for being a singular point
  • $\cos (\pi z)$ is zero, which follows $\tan (\pi z)$ is a simple pole at that point. $\cos (\theta)$ is zero where $\theta$ is a half-integer multiple of $\pi$.
  • $\sin (\pi z^2)$ is zero, which follows $\cot (\pi z^2)$ is a pole of order 2 at that point. $\sin (\theta)$ is zero where $\theta$ is an integer multiple of $\pi$.
Singular points at $\mathbb{C}$
This function is singular at all points such $\cos(\pi z)=0$ and all points such $\sin(\pi z^2)=0$.
  • Case 1: $z$ is a half-integer. Then $z = k + \frac{1}{2}, k \in \mathbb{Z}$. Then $\pi z$ will be a half-integer multiple of $\pi$. Then $\tan(\pi z)$ will have a simple pole at that point since $\sin(\pi z) \neq 0, \cos(\pi z) = 0$ (Simple pole)
  • Case 2: $z^2$ is an integer, $k$. Then $z = \sqrt{k}, k \in \mathbb{Z}$, and $z$ is either on the real or imaginary axis. Then $\pi z^2$ will be an integer multiple of $\pi$, so $\cot^2(\pi z^2)$ will have a pole up to order 2 at that point since $\sin(\pi z^2) = 0$ at that denominator.
    • Case 2a: $z^2$ is a negative integer, $-k$ where $k \in \mathbb{N}$. Then $z = i\sqrt{k}$, on the imaginary axis. Only $\sin^2(\pi z^2)$ is zero. (Double pole)
    • Case 2b: $z^2$ is a positive integer, $k$, but $z$ is irrational.  Only $\sin^2(\pi z^2)$ is zero. (Double pole)
    • Case 2c: $z$ is a positive integer as well as $z^2$. Both $\sin^2(\pi z^2)$ on denominator and $\sin(\pi z) are zero. (Simple pole)
    • Case 2d: $z$ is zero. Then $z, \sin(\pi z), \sin^2(\pi z^2)$ are all zero. (Removable)
There are no branch points on this function as it does not involve fractional powers and logarithms which are known to be multivalued.

Proofs:
Case 1: $z$ is a half-integer. It follows $z$ is real, and is one half more than an integer.
Then $\pi z$ is a half-integer multiple of $\pi$, so $\cos(\pi z) = 0, \sin^2(\pi z) \neq 0$.
Let $z = k + \frac{1}{2}$ such $k \in \mathbb{Z}$. Then $z^2 = (k + \frac{1}{2})^2 = [k^2 + 1] + \frac{1}{4}$, so the square of a half-integer is one quarter greater than an integer. Hence $z^2$ is neither an integer nor a half-integer multiple of $\pi$, so $\cot^2(\pi z^2) \neq 0$.

Then all the terms on the numerator are nonzero, and $\cos(\pi z) \sin(\pi z^2) = 0$ due to the zero value of the cosine. Then $\tan(\pi z)$ has a simple pole and $\csc^2 (\pi z^2)$ is nonzero.

Therefore, for all half-integer $z$, $f$ has a simple pole.

Case 2a: $z$ is an irrational real multiple of $\pi$, but $z^2$ is a positive integer.
Then $\pi z^2$ is an integer multiple of $\pi$, so $\sin^2(\pi z^2) = 0$. Also, $\pi z$ is not a rational multiple of $\pi$, so $\sin(\pi z) \neq 0, \cos(\pi z) \neq 0$. Also, note $\cos^2(\pi z^2) \neq 0$ as the argument of $\cos^2$ is an integer multiple of $\pi$, rather than a half integer multiple.

It follows $\cot^2 (\pi z^2)$ is singular, whereas $z \tan (\pi z)$ is defined. As $\cot (\pi z^2)$ has a simple pole here, it follows $\cot^2 (\pi z^2)$ has a pole of order 2.

Therefore, for all $z \notin \mathbb{Q}$ and $z^2 \in \mathbb{Z}$, $\sin^2(\pi z^2) = 0$. Then $f$ has a pole of order 2 at all points where $z^2$ is a positive integer, but the magnitude of $z$ is irrational.

Case 2b: $z$ is a real multiple of $i\pi$, and $z^2$ is a negative integer.
Then $\pi z^2$ is a negative integer multiple of $\pi$, so $\sin^2(\pi z^2) = 0$, but $\pi z$ is not a rational multiple of $\pi$, so the proof is similar to Case 2a.

Therefore, for all $z$ such $z^2$ is a negative integer, $\sin^2(\pi z^2) = 0$. Then $f$ has a pole of order 2 at all points where $z^2$ is a negative integer, but the magnitude of $z$ is irrational.

Case 2c: $z$ is a nonzero integer multiple of $\pi$.
Then $z^2$ is a positive integer multiple of $\pi$. Then $\cos(\pi z) \neq 0, \cos^2(\pi z^2) \neq 0$, but both $\sin(\pi z) = 0, \sin^2(\pi z^2) = 0$, so we have $f = z^3\frac{\cos^2(\pi z^2)}{\cos(\pi z)} \frac{\sin(\pi z)}{\sin^2(\pi z^2)}$, where $z \frac{\cos^2(\pi z^2)}{\cos(\pi z)}$ is a defined here.

Then $g(z) = \frac{\sin(\pi z)}{\sin^2(\pi z^2)}$ is singular where $z$ is a nonzero integer multiple of $\pi$, and that $f = z \frac{\cos^2(\pi z^2)}{\cos(\pi z)}g$.

$g(z)$ has a simple pole at nonzero integer multiples of $z$.

Therefore, $f$ has a simple pole at all nonzero integer multiples of $z$.

Case 2d: $z$ is zero.
Then $z^2$ is also an integer multiple of $\pi$, so $\cos(\pi z) \neq 0, \cos^2(\pi z^2) \neq 0$. We examine $h(z) = z \frac{\sin(\pi z)}{\sin^2(\pi z^2)}$. such $f = \frac{\cos^2(\pi z^2)}{\cos(\pi z)}h$.

We use Cauchy's Theorem to determine that $h$ and so $f$ is actually analytic on an arbitrarily small disk $D_{\delta,0} = \{z : |z| < \delta\}$.

Therefore, $f$ has a removable singularity at $z = 0$, and the limit approaches 1.

Singular points at $\infty$
Consider $\displaystyle g(z) = f(1/z) = z^{-3} \tan(\pi/z) \cot^2(\pi/z^2)$

This is an essential singularity, as $g$ approaches no limit as $z \mapsto 0$. For example, if $z$ is real and approaches positive infinity, $g$ becomes zero at points that are integer values of $\pi$ and undefined at points $z^2$ is an integer multiple of $\pi$ and nonzero elsewhere. Then it follows $f$ approaches no limit as $z \mapsto \infty$ and will contain infinitely many poles along that real axis. Hence $f(z)$ must have an essential nonisolated singularity at infinity.
« Last Edit: November 04, 2018, 09:26:29 PM by oighea »

Victor Ivrii

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Re: Term Test 2 sample P3
« Reply #2 on: November 04, 2018, 03:25:09 PM »
Try to write a simple listing

About infinity: is the singularity at infinity isolated?

Suspicious points where either $\sin(\pi z^2)=0 \iff z^2\in \mathbb{Z}$ or $\cos(\pi z)=0\iff z\in \frac{1}{2}+\mathbb{Z}$. Those are mutually exclusive.

1. Case $z=\pm \sqrt{n}, n=1,2,\ldots$ and $n$ is not a perfect square OR  $z=\pm \sqrt{n}i, n=1,2,\ldots$ ; then only $\sin (\pi z^2)$ vanishes and it has simple zero; so we get a double pole (which means pole of order $2$).

2. Case $z=\pm m, m=1,2,\ldots$. In this case both $\sin (\pi z^2)$ and $\sin (\pi z)$ have simple zeroes; so we get a simple pole (which means pole of order $1$).

3. Case $z= n+\frac{1}{2}, n\in \mathbb{Z}$ ; then only $\cos (\pi z)$ vanishes and it has simple zero; so we get a simple pole.

4. Finally, $z=0$. Indeed, removable .

5. Infinity: not isolated (from above) and thus does not fall into one of three categories removable, pole or essential singularity.
« Last Edit: December 01, 2018, 08:48:39 AM by Victor Ivrii »