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.
