Find the singularities of $f(z) =\frac{1}{(2\sin z - 1)^2}$. I am just learning about singularities and I was wondering if someone could give me feedback on my work. So I think, for this function, that there are singularities at $z=\frac{\pi}{6}+2k\pi,\frac{5\pi}{6}+2k\pi$, would this be correct? Additionally, I am classifying them as essential and nonremovable. Is this also correct?
To determine that they were nonremovable, I tool the limit of the function approaching the singularities and found that they tended towards infinity.
This $f$ has a singularity anywhere \begin{align*} 2 \sin z - 1 &= 0 \text{,} \\ \sin z &= 1/2 \text{.} \end{align*} This has infinitely many solutions in the reals, \begin{align*} z \in &\{ \arcsin(1/2) + 2 \pi k : k \in \Bbb{Z} \} \cup \\ \quad &\{\pi - \arcsin(1/2) + 2 \pi k : k \in \Bbb{Z}\} \\ = & \{ \pi/6 + 2 \pi k : k \in \Bbb{Z} \} \cup \\ \quad &\{5\pi/6 + 2 \pi k : k \in \Bbb{Z}\} \text{.} \end{align*}
It turns out these are the only two families of solutions, but you have done nothing to show that there are not more solutions. To do so, start with, for $x,y\in\Bbb{R}$, $$ \sin(x+\mathrm{i} \, y) = \sin(x) \cosh(y) + \mathrm{i} \cos(x) \sinh(y) \text{.} $$ Then talk about the zeroes of the real-valued cosine and hyperbolic sine (since $1/2$ has imaginary part $0$), which will restrict $y$ so that $\cosh = 1$, forcing the real solutions to be the only solutions.
If you have shown that the limits are infinity approaching each of these infinitely many singularities, then you have shown that they are not removable. I'm not convinced you have understood the definition/nature of essential singularities.