224. Write the number $z=-2+2 i \sqrt{3}$ in trigonometric form.
Solution. $1^{0}$. Here $a=-2, b=2 \sqrt{3}$. Therefore,

$$
r=\sqrt{a^{2}+b^{2}}=\sqrt{(-2)^{2}+(2 \sqrt{3})^{2}}=\sqrt{4+12}=\sqrt{16}=4
$$

$2^{0}$. Let's represent the number $z$ geometrically (Fig. 11). We see that the number $z$ corresponds to the point $Z$, lying in the II quadrant, and the vector $\overrightarrow{\boldsymbol{z}}$.

$3^{0}$. We find

$$
\cos \varphi=a / r=-2 / 4=-1 / 2 ; \sin \varphi=b / r=2 \sqrt{3} / 4=\sqrt{3} / 2
$$

These relations correspond to the angle $\varphi=180^{\circ}-60^{\circ}=120^{\circ}$ or $\varphi=$ $=\pi-\frac{\pi}{3}=\frac{2 \pi}{3}$.

$4^{0}$. Let's write the given number in trigonometric form:

$$
z=-2+2 i \sqrt{3}=4\left(\cos 120^{\circ}+i \sin 120^{\circ}\right)
$$

or

$$
z=-2+2 i \sqrt{3}=4\left(\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right)
$$