Consider the complex-valued function $\frac{1}{(z^2+1)(z^2+z+1)(z^2-z+1)}$

Question

I have found all the singularities of this function. They are

$\frac{\sqrt 3}{2} + \frac{1}{2}i$, $i$, $-\frac{\sqrt 3}{2} + \frac{1}{2}i$, $-\frac{\sqrt 3}{2} - \frac{1}{2}i$, $-i$, $\frac{\sqrt 3}{2} - \frac{1}{2}i$

and $\frac{1}{2} + \frac{\sqrt 3}{2}i$, $\frac{1}{2} - \frac{\sqrt 3}{2}i$, $-\frac{1}{2} + \frac{\sqrt 3}{2}i$, and $-\frac{1}{2} - \frac{\sqrt 3}{2}i$

However, I am not sure which of all of these is/are a removable singularities.

I am more familiar with poles, but less familiar with the essential and removable discontinuities.

Thank you,

Bayerischer

Answer 1

all of them are removable because they have a finite order. singularities such as $f(z)=e^{\frac{1}{z}}$ at $z=0$ are not removable.