Determine the nature of the singularity of $\frac{z^{3}}{\cos{(z^{5})} - 1}$ at $0$. How can I find if the singularity of $f(z) = \frac{z^{3}}{\cos{z^{5}} - 1}$ at $0$ is essential, removable, or a pole?
I have considered the the Laurent series, and it appears that there are an infinte number of negative powers, so I was inclined to say that this is an essential singularity. However, the solutions say that $f(z)$ has a pole of order $7$ at $0$, by considering zeros of the numerator and denominator.
I am unsure how zeros of the numerator and denominator are related to poles?
Many thanks

$$cos (z^5) = 1 - \frac{(z^5)^2}{2!} + \frac{(z^5)^4}{4!} + ....$$
So, $$cos(z^5)-1 = (1 -\frac{(z^5)^2}{2!} + \frac{(z^5)^4}{4!} + .... - 1) = -\frac{(z^5)^2}{2!} +\frac{(z^5)^4}{4!} - ...$$
Now, $$\frac{z^3}{cos(z^5)-1} = \frac{z^3}{-\frac{(z^5)^2}{2!} +\frac{(z^5)^4}{4!} - ...} = \frac{1}{-\frac{z^7}{2!} +\frac{(z^{17})}{4!} - ...} = \frac{1}{(z^7)(-\frac{1}{2!} +\frac{(z^{10})}{4!} - ...)}$$
By having a look at the denominator, we can say that the pole is of order 7 at z = 0
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