I am doing problem 2.44 from the textbook Optics, 5th edition, by Hecht:
Working with exponentials directly, show that the magnitude of $\psi = Ae^{i \omega t}$ is A. Then rederive the same result using Euler's formula. Prove that $e^{i \alpha}e^{i \beta} = e^{i(\alpha + \beta)}$.
My work is as follows:
$$\begin{align} ||\psi|| &= ||Ae^{i \omega t}|| \\ &= ||A|| \cdot ||e^{i \omega t}|| \ \ \text{(Since $A \in \mathbb{R}$.)} \\ &= A \cdot ||e^{i \omega t}|| \ \ \text{(Since $A$ is the amplitude of the wave, which must be positive.)} \\ &= A(1) \ \ \text{(Since $|z| = |re^{i \omega t}| = r$, which in this case is just equal to 1.)} \\ &= A \end{align}$$
$$\begin{align} ||\psi|| &= || A[ \cos(\omega t) + i \sin(\omega t)] || \\ &= || A || \cdot ||\cos(\omega t) + i \sin(\omega t) || \\ &= || A || \cdot \sqrt{\cos^2(\omega t) + \sin^2(\omega t)} \ \ \text{(By definition of the magnitude of a complex number.)} \\ &= || A || \\ &= A \end{align}$$
$$\begin{align} e^{i \alpha} e^{i \beta} &= [\cos(\alpha) + i \sin(\alpha)][\cos(\beta) + i\sin(\beta)] \\ &= \cos(\alpha) \cos(\beta) + i \sin(\beta) \cos(\alpha) + i \cos(\beta) \sin(\alpha) - \sin(\alpha)\sin(\beta) \\ &= \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta) + i[ \cos(\beta) \sin(\alpha) + \sin(\beta) \cos(\alpha) ] \\ &= \cos(\alpha + \beta) + i[\sin(\alpha + \beta)] \\ &= e^{i(\alpha + \beta)} \end{align}$$
$$\begin{align} e^{i(\alpha + \beta)} &= \cos(\alpha + \beta) + i \sin(\alpha + \beta) \\ &= \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta) + i[ \cos(\beta) \sin(\alpha) + \sin(\beta) \cos(\alpha) ] \\ &= \cos(\alpha) \cos(\beta) + i \sin(\beta) \cos(\alpha) + i \cos(\beta) \sin(\alpha) - \sin(\alpha) \sin(\beta) \\ &= [ \cos(\alpha) + i \sin(\alpha) ] [\cos(\beta) + i \sin(\beta) ] \\ &= e^{i \alpha} e^{i \beta} \end{align}$$
I would greatly appreciate it if people could please review my work for correctness.