$f(z) = \frac{z^2}{1 − \cos z}$ at $z = 0$. What is the easiest way to determined that $z=0$ is a removable singularity?

Question

$f$ is a complex valued function. $f(z) = \frac{z^2}{1 − \cos z}$ at $z = 0$.

What is the easiest way to determined that $z=0$ is a removable singularity.

Is this correct justifation?

The numerator and denominator both have a zero of order two at $z=0$; therefore, $z=0$ is a removable singularity.

Answer 1

Cute but probably not considered simple:

If you want to avoid power series arguments, you could note that the reciprocal of $f(2z)$ is (up to a constant) the square of the expression $g(z)= \frac{\sin z}{z}=\int_{t=0}^{t=1} \cos (tz) dt$ which satisfies $g(0)=1$ and is evidently analytic in $z$ (by differentiation across integral sign).