Determine the nature of the singularity of $\frac{z^{3}}{\cos{(z^{5})} - 1}$ at $0$.

Question

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

Answer 1

It is a pole. Note that\begin{align}\frac{z^3}{\cos(z^5)-1}&=\frac{z^3}{-\frac{(z^5)^2}{2!}+\frac{(z^5)^4}{4!}-\cdots}\\&=\frac1{-\frac{z^7}{2!}+\frac{z^{17}}{4!}-\cdots}\\&=\frac1{z^7}\cdot\frac1{-\frac1{2!}+\frac{z^{10}}{4!}-\cdots}\end{align}So, that function has a pole of order $7$ at $0$.

Answer 2

$$cos (z^5) = 1 - \frac{(z^5)^2}{2!} + \frac{(z^5)^4}{4!} + ....$$

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 .

Answer 3

$\cos z^{5}-1=z^{10} g(z)$ for some holomorphic function $g$ with $g(0) \neq 0$. You can see this from the power series for cosine function. Now $\frac 1 {g(z)}$ is holomorphic in a neighborhood of $0$ so we can expand it in a power series $\sum a_n z^{n}$ with $a_0 \neq 0$. Hence the given function can be expanded as $\frac 1 {z^{7}} \sum a_n z^{n}$. This implies that the function has a pole of order $7$ at $0$.

Answer 4

Just use

• $\lim_{z\to 0}\frac{1-\cos z}{z^2} = \frac{1}{2}$

Hence,

$$\frac{z^{3}}{\cos{(z^{5})} - 1} = \underbrace{\frac{1}{\frac{\cos{(z^{5})} - 1}{z^{10}}}}_{\stackrel{z \to 0}{\longrightarrow}-2}\cdot\frac{1}{z^7} \Rightarrow \mbox{ pole of order }7$$