235. Write the number $z=3-3 i \sqrt{3}$ in trigonometric and exponential forms.
Solution. Since $a=3, b=-3 \sqrt{3}$,

then $r=\sqrt{a^{2}+b^{2}}=\sqrt{3^{2}+(-3 \sqrt{3})^{2}}=$ $=\sqrt{9+9 \cdot 3}=6$. Geometrically, we determine that the number $z$ corresponds to the point $Z$, which lies in the IV quadrant (Fig. 13). We form the ratios

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-104.jpg?height=361&width=387&top_left_y=1458&top_left_x=768)

Fig. 13

$$
\cos \varphi=a / r=3 / 6=1 / 2, \sin \varphi=b / r=-3 \sqrt{3} / 6=-\sqrt{3} / 2
$$

From this, it follows that $\varphi=360^{\circ}-60^{\circ}=300^{\circ}$ or $\varphi=2 \pi-\frac{\pi}{3}=\frac{5 \pi}{3}$. Therefore, $r=6, \varphi=\frac{5 \pi}{3}$. Thus, $z=6\left(\cos \frac{5 \pi}{3}+i \sin \frac{5 \pi}{3}\right)$ is the trigonometric form, and $z=6 e^{\frac{5 \pi}{3} i}$ is the exponential form of the number.