Complex Analysis - Proof that $1-z^{2}$ is analytic

Question

You may find it simple but I don t really see how I can prove that :

$$f(z) = 1-z^{2}$$

where $f$ is define on : $\mathbb{C} \setminus [- \infty ;-1 ]\cup[1;\infty] $

(Bonus : Can we say that $f(z) \in \mathbb{C} \setminus \left]\infty;0\right] \text{?} ) $

Thanks in advance for your help guys.

Answer 1

If it's differentiable it's analytic. The derivative is $-2z$, defined everywhere.