# Find all singularities of$\ \frac{\cos z - \cos (2z)}{z^4} \$

How do I find all singularities of$\ \frac{\cos z - \cos (2z)}{z^4} \$

It seems like there is only one (z = 0)?

How do I decide if it is isolated or nonisolated?

And if it is isolated, how do I decide if it is removable or not removable?

If it is non isolated, how do I decide the orders of the singularities?

Thanks!!!

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$0$ is the only singularity of that function. Do you know what it means for a singularity to be isolated? Do you know what it means for a singularity to be removable? Do you know what the order of a singularity is? –  anonymous Nov 28 '12 at 5:50

$$\cos z = 1-\frac{z^2}{2}+\cdots$$ $$\cos2z=1-\frac{(2z)^2}{2}+\cdots$$ $$\frac{\cos z-\cos2z}{z^4}= \frac{3}{2z^2}+\left(\frac{-15}{4!}+a_1z^2 +\cdots\right),$$ hence at $z=0$ there is a pole .