Упр.17.7 ГДЗ Никольский Потапов 11 класс (Алгебра)

Задача

Выполните умножение комплексных чисел.

Подробный ответ

Используем формулу умножения комплексных чисел, записанных в тригонометрической форме:

$$\left(\cos \alpha + i \sin \alpha\right)\left(\cos \beta + i \sin \beta\right) = \cos(\alpha+\beta)+i\sin(\alpha+\beta).$$

а)

$$\left(\cos \frac{5\pi}{3}+i\sin \frac{5\pi}{3}\right)\left(\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}\right) = \cos \left(\frac{5\pi}{3}+\frac{\pi}{3}\right)+i\sin \left(\frac{5\pi}{3}+\frac{\pi}{3}\right)$$
$$= \cos 2\pi+i\sin 2\pi=1.$$

б)

$$6\left(\cos \frac{5\pi}{4}+i\sin \frac{5\pi}{4}\right)\left(\cos \frac{7\pi}{4}+i\sin \frac{7\pi}{4}\right)$$
$$=6\left(\cos \left(\frac{5\pi}{4}+\frac{7\pi}{4}\right)+i\sin \left(\frac{5\pi}{4}+\frac{7\pi}{4}\right)\right)$$
$$=6(\cos 3\pi+i\sin 3\pi)=-6.$$

в)

$$8\left(\cos \frac{2\pi}{3}+i\sin \frac{2\pi}{3}\right)\left(\cos \frac{5\pi}{4}+i\sin \frac{5\pi}{4}\right)$$
$$=8\left(\cos \left(\frac{2\pi}{3}+\frac{5\pi}{4}\right)+i\sin \left(\frac{2\pi}{3}+\frac{5\pi}{4}\right)\right)$$
$$=8\left(\cos \frac{23\pi}{12}+i\sin \frac{23\pi}{12}\right) =8\left(\cos \left(2\pi-\frac{\pi}{12}\right)+i\sin \left(2\pi-\frac{\pi}{12}\right)\right)$$
$$=8\left(\cos \frac{\pi}{12}-i\sin \frac{\pi}{12}\right).$$

г)

$$21\left(\cos \frac{7\pi}{6}+i\sin \frac{7\pi}{6}\right)\left(\cos \frac{7\pi}{12}+i\sin \frac{7\pi}{12}\right)$$
$$=21\left(\cos \left(\frac{7\pi}{6}+\frac{7\pi}{12}\right)+i\sin \left(\frac{7\pi}{6}+\frac{7\pi}{12}\right)\right)$$
$$=21\left(\cos \frac{7\pi}{4}+i\sin \frac{7\pi}{4}\right) =21\left(\cos \left(2\pi-\frac{\pi}{4}\right)+i\sin \left(2\pi-\frac{\pi}{4}\right)\right)$$
$$=21\left(\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\right) =10{,}5\sqrt{2}(1-i).$$