Упр.36.14 ГДЗ Мордковича 10 класс профильный уровень (Алгебра)

1. а) Вычислите $$z^{12}$$, если $$z=2\cos\frac{\pi}{8}\left(\sin\frac{3\pi}{4}+i+i\cos\frac{3\pi}{4}\right)$$; б) вычислите $$z^{30}$$, если $$z=2\sin\frac{\pi}{12}\left(1-\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right)$$.

а) Преобразуем число $$z$$:

$$z=2\cos\frac{\pi}{8}\left(\sin\frac{3\pi}{4}+i+i\cos\frac{3\pi}{4}\right)$$

$$z=2\cos\frac{\pi}{8}\left(2\sin\frac{3\pi}{8}\cos\frac{3\pi}{8}+2i\cos^2\frac{3\pi}{8}\right)$$

$$z=4\cos\frac{\pi}{8}\cos\frac{3\pi}{8}\left(\sin\frac{3\pi}{8}+i\cos\frac{3\pi}{8}\right)$$

$$z=4\cos\frac{\pi}{8}\cos\frac{3\pi}{8}\left(\cos\frac{\pi}{8}+i\sin\frac{\pi}{8}\right)$$

Так как $$\cos\frac{3\pi}{8}=\sin\frac{\pi}{8}$$, получаем:

$$z=4\cos\frac{\pi}{8}\sin\frac{\pi}{8}\left(\cos\frac{\pi}{8}+i\sin\frac{\pi}{8}\right) =2\sin\frac{\pi}{4}\left(\cos\frac{\pi}{8}+i\sin\frac{\pi}{8}\right)$$

$$z=\sqrt{2}\left(\cos\frac{\pi}{8}+i\sin\frac{\pi}{8}\right)$$

Тогда по формуле Муавра:

$$z^{12}=\left(\sqrt{2}\right)^{12}\left(\cos\frac{12\pi}{8}+i\sin\frac{12\pi}{8}\right) =2^6\left(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2}\right)$$

$$z^{12}=64(0-i)=-64i$$

б) Преобразуем число $$z$$:

$$z=2\sin\frac{\pi}{12}\left(1-\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right)$$

$$z=2\sin\frac{\pi}{12}\left(2\sin^2\frac{5\pi}{12}+2i\sin\frac{5\pi}{12}\cos\frac{5\pi}{12}\right)$$

$$z=4\sin\frac{\pi}{12}\sin\frac{5\pi}{12}\left(\sin\frac{5\pi}{12}+i\cos\frac{5\pi}{12}\right)$$

$$z=4\sin\frac{\pi}{12}\cos\frac{\pi}{12}\left(\cos\frac{\pi}{12}+i\sin\frac{\pi}{12}\right)$$

Так как $$2\sin\frac{\pi}{12}\cos\frac{\pi}{12}=\sin\frac{\pi}{6}$$, имеем:

$$z=2\sin\frac{\pi}{6}\left(\cos\frac{\pi}{12}+i\sin\frac{\pi}{12}\right) =\cos\frac{\pi}{12}+i\sin\frac{\pi}{12}$$

Тогда

$$z^{30}=\left(\cos\frac{\pi}{12}+i\sin\frac{\pi}{12}\right)^{30} =\cos\frac{30\pi}{12}+i\sin\frac{30\pi}{12}$$

$$z^{30}=\cos\frac{5\pi}{2}+i\sin\frac{5\pi}{2}=i$$

Ответ

а) $$-64i$$; б) $$i$$.