Is it a removable singularity?

Question

In the function:

$$ f(z)=2iz\frac{(1-z^{2})^{\frac{1}{2}}}{1-2z^{2}} \qquad \qquad (z \in \mathbb{Z}) \,\, , $$

There is a singularity at the point $z=\pm \sqrt{1/2}$. Is that a removable singularity? How to know this?

Answer 1

Is that a removable singularity? How to know this?

No, this is not a removable singularity : the numerator does not vanish at $z=1/\sqrt{2}$.

Answer 2

No, numerator gives zeros of $0,\pm 1$ and denominator gives zeros of $\displaystyle \pm \frac{1}{\sqrt{2}}$ which are the poles of $f(z)$. Unlike $\displaystyle \frac{\sin x}{x}$, both the numerator and denominator have the same "zero", which is $z=0$, of order one.