### Author Topic: Term Test 2 sample P3  (Read 1972 times)

#### Victor Ivrii ##### 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 ##### 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)$must have an essential nonisolated singularity at infinity. « Last Edit: November 04, 2018, 09:26:29 PM by oighea » #### Victor Ivrii ##### Re: Term Test 2 sample P3 « Reply #2 on: November 04, 2018, 03:25:09 PM » Try to write a simple listing About infinity: is the singularity at infinity isolated? Suspicious points where either$\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 »