Power series of complex cosine

Statement

For each $z \in \mathbb{C}$, the series $\sum_{n \ge 0} (-1)^n z^{2n} / (2n)!$ converges absolutely, and $$\cos z = \sum_{n=0}^{\infty} (-1)^n \frac{z^{2n}}{(2n)!} = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \cdots,$$ where $\cos$ is the complex cosine.

Remarks

Proof: put $iz$ and $-iz$ in the series of the complex exponential and add. The terms with odd powers cancel, and $i^{2n} = (-1)^n$. In the same way, $\sin z = \sum_{n \ge 0} (-1)^n z^{2n+1} / (2n+1)!$ for each $z \in \mathbb{C}$. The series shows that $\cos$ is an even function: $\cos(-z) = \cos z$.

Explanations

L1 human

The series of $\cos$ has only the even powers of $e^z$'s series, with signs that alternate. For example, $\cos(0.1) \approx 1 - 0.005 = 0.995$ from the first two terms.