Упр.15.9 ГДЗ Мерзляк 11 класс Углубленный уровень (Алгебра)

1) z=3v3(cos(2п/3)+isin(2п/3)), n=3; 3) z=-32i, n=5;
2) z=8(cos(6/7)+isin(6/7)), n=3; 4) z=v3+i, n=4.

1. $$z=3\sqrt{3}\left(\cos \frac{2\pi}{3}+i\sin \frac{2\pi}{3}\right), \quad n=3.$$

   По формуле корней $$n$$-й степени:

   $$
   \sqrt[3]{z_k}=\sqrt[3]{27}\left(\cos \frac{\frac{2\pi}{3}+2\pi k}{3}+i\sin \frac{\frac{2\pi}{3}+2\pi k}{3}\right), \quad k=0,1,2.
   $$

   Так как $$\sqrt[3]{27}=3$$, получаем:

   $$
   \sqrt[3]{z_1}=3\left(\cos \frac{2\pi}{9}+i\sin \frac{2\pi}{9}\right),
   $$

   $$
   \sqrt[3]{z_2}=3\left(\cos \frac{8\pi}{9}+i\sin \frac{8\pi}{9}\right),
   $$

   $$
   \sqrt[3]{z_3}=3\left(\cos \frac{14\pi}{9}+i\sin \frac{14\pi}{9}\right).
   $$

2. $$z=8\left(\cos \frac{6}{7}+i\sin \frac{6}{7}\right), \quad n=3.$$

   Тогда

   $$
   \sqrt[3]{z_k}=2\left(\cos \frac{\frac{6}{7}+2\pi k}{3}+i\sin \frac{\frac{6}{7}+2\pi k}{3}\right), \quad k=0,1,2.
   $$

   Следовательно,

   $$
   \sqrt[3]{z_1}=2\left(\cos \frac{2}{7}+i\sin \frac{2}{7}\right),
   $$

   $$
   \sqrt[3]{z_2}=2\left(\cos \left(\frac{2}{7}+\frac{2\pi}{3}\right)+i\sin \left(\frac{2}{7}+\frac{2\pi}{3}\right)\right),
   $$

   $$
   \sqrt[3]{z_3}=2\left(\cos \left(\frac{2}{7}+\frac{4\pi}{3}\right)+i\sin \left(\frac{2}{7}+\frac{4\pi}{3}\right)\right).
   $$

3. $$z=-32i, \quad n=5.$$

   Найдём модуль и аргумент числа:

   $$r=\sqrt{0^2+(-32)^2}=32,$$

   $$\cos \varphi=0,\quad \sin \varphi=-1,$$

   значит,

   $$z=32\left(\cos \left(-\frac{\pi}{2}\right)+i\sin \left(-\frac{\pi}{2}\right)\right).$$

   Тогда корни пятой степени:

   $$
   \sqrt[5]{z_k}=2\left(\cos \frac{-\frac{\pi}{2}+2\pi k}{5}+i\sin \frac{-\frac{\pi}{2}+2\pi k}{5}\right), \quad k=0,1,2,3,4.
   $$

   Получаем:

   $$
   \sqrt[5]{z_1}=2\left(\cos \left(-\frac{\pi}{10}\right)+i\sin \left(-\frac{\pi}{10}\right)\right),
   $$

   $$
   \sqrt[5]{z_2}=2\left(\cos \frac{3\pi}{10}+i\sin \frac{3\pi}{10}\right),
   $$

   $$
   \sqrt[5]{z_3}=2\left(\cos \frac{7\pi}{10}+i\sin \frac{7\pi}{10}\right),
   $$

   $$
   \sqrt[5]{z_4}=2\left(\cos \frac{11\pi}{10}+i\sin \frac{11\pi}{10}\right),
   $$

   $$
   \sqrt[5]{z_5}=2\left(\cos \frac{3\pi}{2}+i\sin \frac{3\pi}{2}\right)=-2i.
   $$

4. $$z=\sqrt{3}+i, \quad n=4.$$

   Найдём модуль и аргумент:

   $$r=\sqrt{(\sqrt{3})^2+1^2}=\sqrt{4}=2,$$

   $$\cos \varphi=\frac{\sqrt{3}}{2},\quad \sin \varphi=\frac{1}{2},$$

   значит,

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

   Тогда

   $$
   \sqrt[4]{z_k}=\sqrt[4]{2}\left(\cos \frac{\frac{\pi}{6}+2\pi k}{4}+i\sin \frac{\frac{\pi}{6}+2\pi k}{4}\right), \quad k=0,1,2,3.
   $$

   Следовательно,

   $$
   \sqrt[4]{z_1}=\sqrt[4]{2}\left(\cos \frac{\pi}{24}+i\sin \frac{\pi}{24}\right),
   $$

   $$
   \sqrt[4]{z_2}=\sqrt[4]{2}\left(\cos \frac{13\pi}{24}+i\sin \frac{13\pi}{24}\right),
   $$

   $$
   \sqrt[4]{z_3}=\sqrt[4]{2}\left(\cos \frac{25\pi}{24}+i\sin \frac{25\pi}{24}\right),
   $$

   $$
   \sqrt[4]{z_4}=\sqrt[4]{2}\left(\cos \frac{37\pi}{24}+i\sin \frac{37\pi}{24}\right).
   $$

Ответ

1) $$3\left(\cos \frac{2\pi}{9}+i\sin \frac{2\pi}{9}\right),\ 3\left(\cos \frac{8\pi}{9}+i\sin \frac{8\pi}{9}\right),\ 3\left(\cos \frac{14\pi}{9}+i\sin \frac{14\pi}{9}\right)$$;

2) $$2\left(\cos \frac{2}{7}+i\sin \frac{2}{7}\right),\ 2\left(\cos \left(\frac{2}{7}+\frac{2\pi}{3}\right)+i\sin \left(\frac{2}{7}+\frac{2\pi}{3}\right)\right),\ 2\left(\cos \left(\frac{2}{7}+\frac{4\pi}{3}\right)+i\sin \left(\frac{2}{7}+\frac{4\pi}{3}\right)\right)$$;

3) $$2\left(\cos \left(-\frac{\pi}{10}\right)+i\sin \left(-\frac{\pi}{10}\right)\right),\ 2\left(\cos \frac{3\pi}{10}+i\sin \frac{3\pi}{10}\right),\ 2\left(\cos \frac{7\pi}{10}+i\sin \frac{7\pi}{10}\right),\ 2\left(\cos \frac{11\pi}{10}+i\sin \frac{11\pi}{10}\right),\ -2i$$;

4) $$\sqrt[4]{2}\left(\cos \frac{\pi}{24}+i\sin \frac{\pi}{24}\right),\ \sqrt[4]{2}\left(\cos \frac{13\pi}{24}+i\sin \frac{13\pi}{24}\right),\ \sqrt[4]{2}\left(\cos \frac{25\pi}{24}+i\sin \frac{25\pi}{24}\right),\ \sqrt[4]{2}\left(\cos \frac{37\pi}{24}+i\sin \frac{37\pi}{24}\right)$$.