Classifying singularities of a function

[penroseandpaper]

Homework Statement

Classify the singularity for the function

$$ f(z)=\frac{cos(z)} {(z-π/2)^7} $$

Relevant Equations

Theorems and lemmas related to essential singularities, poles and removable singularities.

I came across this question on chegg for practice as I'm self learning complex analysis, but became stumped on it and without access to the solution am unable to check.

Let $$ f(z)=\frac{cos(z)} {(z-π/2)^7} $$. Then the singularity is at π/2. And on first appearance, it looks like a pole of order 7. However, multiplying by $$(z-π/2)^7 $$ and taking the limit evaluates to zero - meaning it isn't a pole of order 7.

Similarly, the limit of f(z) doesn't exist either, so it can't be a removable singularity.

That must mean it is essential. But is that right and how can I go about proving that, other than showing the above are true leaving only the essential singularity as an option.

Thanks,

 

[andrewkirk]

...it can't be a removable singularity.

That must mean it is essential. But is that right ...

No it's not right because not all non-removable singularities are essential. An essential singularity is a singularity that is not a pole of any order. Removable singularities are poles of order 0. So the point $\pi/2$ could be a pole of some finite order. That order cannot be greater than 7 since $g(z):= f(z)(z-\pi/2)^7=\cos z$ is holomorphic.
Since you have shown the pole doesn't have order 7, I suggest you investigate whether it has order 6. If that doesn't work try 5. I'd be surprised if the order were lower than 5.

 

[FactChecker]

Let $$ f(z)=\frac{cos(z)} {(z-π/2)^7} $$. Then the singularity is at π/2. And on first appearance, it looks like a pole of order 7. However, multiplying by $$(z-π/2)^7 $$ and taking the limit evaluates to zero - meaning it isn't a pole of order 7.

Consider the Taylor series expansion of cos(z) at z=$\pi/2$. You know what the derivative of cos(z) is so you should know something about a coefficient in the Taylor series. Use that to determine the order of the pole of f(z).

Similarly, the limit of f(z) doesn't exist either, so it can't be a removable singularity.

There are other possible orders of the pole. A zero of cos(z) of order n will make the pole of f(z) be a pole of order 7-n.

That must mean it is essential. But is that right and how can I go about proving that, other than showing the above are true leaving only the essential singularity as an option.

It is not an essential singularity.