Solution: We will solve this by two methods; one is by using a result (which will be important) and other is by looking at the expansion of the given function around $z=0$. Recall that a point $z_0$ is called a removable singularity of $f$ is \[ \lim_{z \to z_0} f(z) \text{ exists }. \] We note that \begin{align*} & 1 - \cos z = 1 - \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}z^{2n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{(2n)!}z^{2n} \\ \implies & \frac{\cos z}{z^2} = \frac{1}{z^2} \sum_{n = 1}^{\infty} \frac{(-1)^n}{(2n)!}z^{2n} = \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(2n+2)!}z^{2n}. \end{align*} Note that the function \[ g(z) = \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(2n+2)!}z^{2n} = \frac{\cos z}{z^2}, \] and as \[ \lim_{z \to 0} g(z) = -\frac{1}{2}, \] the point $z=0$ is a removable singularity.

We recall the following theorem which will be important to verify if some singularity is removable or not.

Using the above theorem, we need to see if the limit of $z \frac{\cos z}{z^2}$ vanishes or not as $z$ approaches to zero. Observe that \begin{align*} \lim_{z \to 0} z\cdot \frac{\cos z}{z^2} = \lim_{z \to 0} \frac{\cos z}{z} = 0. \end{align*} Therefore, $z=0$ is a removable singularity.