About $e^{i z} = \cos z + i \sin z$ in Michael Spivak "Calculus 3rd Edition". I am reading "Calculus 3rd Edition" by Michael Spivak.
The author wrote as follows (p. 555):

Moreover, if we replace $z$ by $i z$ in the series for $e^z$, and make a rearrangement of the terms (justified by absolute convergence), something particularly interesting happens:
$$e^{i z} = 1 + i z + \frac{(iz)^2}{2!} + \frac{(iz)^3}{3!} + \frac{(iz)^4}{4!} + \frac{(iz)^5}{5!} + \cdots \\
=1 + iz - \frac{z^2}{2!} - \frac{i z^3}{3!} + \frac{z^4}{4!} + \frac{i z^5}{5!} + \cdots \\
= (1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \cdots) + i (z - \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots),$$
so $$e^{i z} = \cos z + i \sin z.$$

But I think the author didn't use a rearrangement of the terms at all.
Am I right?
Let
$$c_n := 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \cdots + (-1)^n \frac{z^{2 n}}{(2 n)!},$$
$$s_n := z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots + (-1)^n \frac{z^{2 n + 1}}{(2 n + 1)!},$$
$$e_n := 1 + i z + \frac{(iz)^2}{2!} + \frac{(iz)^3}{3!} + \frac{(iz)^4}{4!} + \frac{(iz)^5}{5!} + \cdots + \frac{(iz)^n}{n!}.$$
Then, $$e_{2 n + 1} = c_n + i s_n,$$
$$\lim_{n \to \infty} e_{2 n + 1} = e^{i z},$$
$$\lim_{n \to \infty} c_n + i s_n = \lim_{n \to \infty} c_n + i \lim_{n \to \infty} s_n = \cos z + i \sin z,$$
so $$e^{i z} = \cos z + i \sin z.$$

Your statement:

But I think the author didn't use a rearrangement of the terms at all.

He did re-arrange the terms as said...See:

