Example 3. Find the modulus and argument of the complex number

$$
z=-\sin \frac{\pi}{8}-i \cos \frac{\pi}{8}
$$
Solution. We have

$$
x=-\sin \frac{\pi}{8}<0, \quad y=-\cos \frac{\pi}{8}<0
$$

The principal value of the argument according to (1) will be

$$
\begin{aligned}
\arg z & =-\pi+\operatorname{arctg}\left(\operatorname{ctg} \frac{\pi}{8}\right)=-\pi+\operatorname{arctg}\left[\operatorname{tg}\left(\frac{\pi}{2}-\frac{\pi}{8}\right)\right]= \\
& =-\pi+\operatorname{arctg}\left(\operatorname{tg} \frac{3}{8} \pi\right)=-\pi+\frac{3}{8} \pi=-\frac{5}{8} \pi
\end{aligned}
$$

Therefore,

$\operatorname{Arg} z=-\frac{5}{8} \pi+2 k \pi \quad(k=0, \pm 1, \pm 2, \ldots), \quad|z|=\sqrt{\sin ^{2} \frac{\pi}{8}+\cos ^{2} \frac{\pi}{8}}=1 . \quad \nabla$

## Problems for independent solution

9. In the following problems, find the modulus and the principal value of the argument of the complex numbers:
a) $z=4+3 i$
b) $z=-2+2 \sqrt{3} i$
c) $z=-7-i$
d) $z=-\cos \frac{\pi}{5}+i \sin \frac{\pi}{5}$

e) $z=4-3 i$;
f) $z=\cos \alpha-i \sin \alpha \quad\left(\pi<\alpha<\frac{3}{2} \pi\right)$

Any complex number $z=x+i y(z \neq 0)$ can be written in trigonometric form

$$
z=\rho(\cos \varphi+i \sin \varphi), \quad \text { where } \quad \rho=|z|, \quad \varphi=\operatorname{Arg} z
$$