Proof of the existence and uniqueness of the principal value of the argument

Proof

We first build an argument in $(-\pi, \pi]$, then show that there is only one. Here $\sin$ and $\cos$ are the real functions given by their power series.

1. Let $z \neq 0$ and let $r = |z|$ be its modulus. Since $\operatorname{Re} z \neq 0$ or $\operatorname{Im} z \neq 0$, we have $r > 0$.
2. Let $a = \operatorname{Re} z / r$ and $b = \operatorname{Im} z / r$. Then $a^2 + b^2 = r^2 / r^2 = 1$. So $a^2 \le 1$, that is, $-1 \le a \le 1$.
3. The function $\cos$ is even and $\sin$ is odd: $\cos(-x) = \cos x$ and $\sin(-x) = -\sin x$ for real $x$. Indeed, $(-x)^{2n} = x^{2n}$ and $(-x)^{2n+1} = -x^{2n+1}$ in each term of the power series.
4. Let $\theta_0 = \arccos a$, the arccosine of $a$. Then $0 \le \theta_0 \le \pi$ and $\cos \theta_0 = a$.
5. By the Pythagorean identity, $\sin^2 \theta_0 = 1 - a^2 = b^2$. So $\sin \theta_0 = b$ or $\sin \theta_0 = -b$.
6. If $\sin \theta_0 = b$, let $\theta_1 = \theta_0$. Otherwise let $\theta_1 = -\theta_0$. By step 3, in both cases $\cos \theta_1 = a$, $\sin \theta_1 = b$ and $-\pi \le \theta_1 \le \pi$.
7. If $\theta_1 = -\pi$, let $\theta = \theta_1 + 2 \pi = \pi$. Otherwise let $\theta = \theta_1$. By periodicity, $\cos \theta = a$ and $\sin \theta = b$, and now $-\pi < \theta \le \pi$.
8. Then $r (\cos \theta + i \sin \theta) = r a + r b \, i = \operatorname{Re} z + (\operatorname{Im} z) i = z$. So $\theta$ is an argument of $z$ in $(-\pi, \pi]$.
9. For uniqueness, we first prove a claim. Let $t$ be real with $\cos t = 1$, $\sin t = 0$ and $-2\pi < t < 2\pi$. Then $t = 0$.
10. Proof of the claim. By the angle addition formulas, $\sin(x + t) = \sin x \cos t + \cos x \sin t = \sin x$ for all real $x$. Also, using step 3, $\sin(x - t) = \sin x \cos t - \cos x \sin t = \sin x$.
11. If $t > 0$, step 10 makes $t$ a positive period of $\sin$ smaller than $2\pi$. If $t < 0$, step 10 makes $-t$ such a period. Both are impossible, because $2\pi$ is the smallest positive period of $\sin$, by periodicity. So $t = 0$.
12. Now let $\theta$ and $\varphi$ be arguments of $z$, both in $(-\pi, \pi]$. Then $r(\cos \theta + i \sin \theta) = r(\cos \varphi + i \sin \varphi)$.
13. Divide by the real number $r > 0$ and compare parts, by equality of complex numbers. This gives $\cos \theta = \cos \varphi$ and $\sin \theta = \sin \varphi$.
14. Let $t = \theta - \varphi$. By the angle addition formulas and step 3, $\cos t = \cos \theta \cos \varphi + \sin \theta \sin \varphi$. By step 13, this is $\cos^2 \varphi + \sin^2 \varphi = 1$.
15. In the same way, $\sin t = \sin \theta \cos \varphi - \cos \theta \sin \varphi = \sin \varphi \cos \varphi - \cos \varphi \sin \varphi = 0$.
16. Both $\theta$ and $\varphi$ lie in $(-\pi, \pi]$, so $-2\pi < t < 2\pi$. By the claim in step 9, $t = 0$, that is, $\theta = \varphi$.
17. So $z$ has exactly one argument in $(-\pi, \pi]$.