Страница 141 номер 15.10, ГДЗ по алгебре за 11 класс к учебнику Мерзляка углубленный уровень. Алгебра и начала математического анализа

Глава 3. Комплексные числа. §15. Умножение и деление комплексных чисел, записанных в тригонометрической форме. Корень n-й степени из комплексного числа. Страница 141. Номер 15.10

Решение:

Корни \(n\)-й степени из числа \(z = r(\cos \varphi + i\sin \varphi) \ne 0\), где \(k \in \{0,\ 1,\ \ldots,\ n - 1\}\): \[\omega_k = \sqrt[n]{r}\left(\cos \frac{\varphi + 2\pi k}{n} + i\sin \frac{\varphi + 2\pi k}{n}\right).\]

1) \(z = 4\left(\cos \frac{16\pi}{19} + i\sin \frac{16\pi}{19}\right)\), \(n = 4\);
\(\sqrt[4]{4} = \sqrt{2}\); угол \(\frac{\frac{16\pi}{19} + 2\pi k}{4} = \frac{4\pi}{19} + \frac{\pi k}{2}\), где \(k \in \{0,\ 1,\ 2,\ 3\}\): \[\begin{aligned} \omega_0 &= \sqrt{2}\left(\cos \frac{4\pi}{19} + i\sin \frac{4\pi}{19}\right), \\ \omega_1 &= \sqrt{2}\left(\cos \frac{27\pi}{38} + i\sin \frac{27\pi}{38}\right), \\ \omega_2 &= \sqrt{2}\left(\cos \frac{23\pi}{19} + i\sin \frac{23\pi}{19}\right), \\ \omega_3 &= \sqrt{2}\left(\cos \frac{65\pi}{38} + i\sin \frac{65\pi}{38}\right). \end{aligned}\] Здесь \(\frac{4\pi}{19} + \frac{\pi}{2} = \frac{27\pi}{38}\), \(\frac{4\pi}{19} + \pi = \frac{23\pi}{19}\), \(\frac{4\pi}{19} + \frac{3\pi}{2} = \frac{65\pi}{38}\).

2) \(z = 125\left(\cos \frac{9}{11} + i\sin \frac{9}{11}\right)\), \(n = 3\);
\(\sqrt[3]{125} = 5\); угол \(\frac{\frac{9}{11} + 2\pi k}{3} = \frac{3}{11} + \frac{2\pi k}{3}\), где \(k \in \{0,\ 1,\ 2\}\): \[\begin{aligned} \omega_0 &= 5\left(\cos \frac{3}{11} + i\sin \frac{3}{11}\right), \\ \omega_1 &= 5\left(\cos\left(\frac{3}{11} + \frac{2\pi}{3}\right) + i\sin\left(\frac{3}{11} + \frac{2\pi}{3}\right)\right), \\ \omega_2 &= 5\left(\cos\left(\frac{3}{11} + \frac{4\pi}{3}\right) + i\sin\left(\frac{3}{11} + \frac{4\pi}{3}\right)\right). \end{aligned}\]

3) \(z = 64\), \(n = 6\);
\(64 = 64(\cos 0 + i\sin 0)\), \(\sqrt[6]{64} = 2\); угол \(\frac{0 + 2\pi k}{6} = \frac{\pi k}{3}\), где \(k \in \{0,\ 1,\ 2,\ 3,\ 4,\ 5\}\): \[\begin{aligned} \omega_0 &= 2(\cos 0 + i\sin 0) = 2, \\ \omega_1 &= 2\left(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}\right) = 2\left(\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) = 1 + \sqrt{3}i, \\ \omega_2 &= 2\left(\cos \frac{2\pi}{3} + i\sin \frac{2\pi}{3}\right) = 2\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) = -1 + \sqrt{3}i, \\ \omega_3 &= 2(\cos \pi + i\sin \pi) = -2, \\ \omega_4 &= 2\left(\cos \frac{4\pi}{3} + i\sin \frac{4\pi}{3}\right) = 2\left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) = -1 - \sqrt{3}i, \\ \omega_5 &= 2\left(\cos \frac{5\pi}{3} + i\sin \frac{5\pi}{3}\right) = 2\left(\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) = 1 - \sqrt{3}i. \end{aligned}\]

4) \(z = \sqrt{3} - i\), \(n = 4\);
\(|z| = \sqrt{3 + 1} = 2\), поэтому \[\sqrt{3} - i = 2\left(\frac{\sqrt{3}}{2} - \frac{1}{2}i\right) = 2\left(\cos\left(-\frac{\pi}{6}\right) + i\sin\left(-\frac{\pi}{6}\right)\right).\] \(\sqrt[4]{r} = \sqrt[4]{2}\); угол \(\frac{-\frac{\pi}{6} + 2\pi k}{4} = -\frac{\pi}{24} + \frac{\pi k}{2}\), где \(k \in \{0,\ 1,\ 2,\ 3\}\): \[\begin{aligned} \omega_0 &= \sqrt[4]{2}\left(\cos\left(-\frac{\pi}{24}\right) + i\sin\left(-\frac{\pi}{24}\right)\right), \\ \omega_1 &= \sqrt[4]{2}\left(\cos \frac{11\pi}{24} + i\sin \frac{11\pi}{24}\right), \\ \omega_2 &= \sqrt[4]{2}\left(\cos \frac{23\pi}{24} + i\sin \frac{23\pi}{24}\right), \\ \omega_3 &= \sqrt[4]{2}\left(\cos \frac{35\pi}{24} + i\sin \frac{35\pi}{24}\right). \end{aligned}\]

Ответ: 1) \(\sqrt{2}\left(\cos \frac{4\pi}{19} + i\sin \frac{4\pi}{19}\right)\); \(\sqrt{2}\left(\cos \frac{27\pi}{38} + i\sin \frac{27\pi}{38}\right)\); \(\sqrt{2}\left(\cos \frac{23\pi}{19} + i\sin \frac{23\pi}{19}\right)\); \(\sqrt{2}\left(\cos \frac{65\pi}{38} + i\sin \frac{65\pi}{38}\right)\); 2) \(5\left(\cos \frac{3}{11} + i\sin \frac{3}{11}\right)\); \(5\left(\cos\left(\frac{3}{11} + \frac{2\pi}{3}\right) + i\sin\left(\frac{3}{11} + \frac{2\pi}{3}\right)\right)\); \(5\left(\cos\left(\frac{3}{11} + \frac{4\pi}{3}\right) + i\sin\left(\frac{3}{11} + \frac{4\pi}{3}\right)\right)\); 3) \(-2\); \(2\); \(1 + \sqrt{3}i\); \(1 - \sqrt{3}i\); \(-1 + \sqrt{3}i\); \(-1 - \sqrt{3}i\); 4) \(\sqrt[4]{2}\left(\cos\left(-\frac{\pi}{24}\right) + i\sin\left(-\frac{\pi}{24}\right)\right)\); \(\sqrt[4]{2}\left(\cos \frac{11\pi}{24} + i\sin \frac{11\pi}{24}\right)\); \(\sqrt[4]{2}\left(\cos \frac{23\pi}{24} + i\sin \frac{23\pi}{24}\right)\); \(\sqrt[4]{2}\left(\cos \frac{35\pi}{24} + i\sin \frac{35\pi}{24}\right)\).

Решение:

Найдём корни \(n\)-й степени из числа \(z\) в каждом пункте. Если \(z = r(\cos \varphi + i\sin \varphi)\) — тригонометрическая форма числа \(z \ne 0\), то все \(n\) корней \(n\)-й степени из него находятся по формуле корней \(n\)-й степени, где \(k \in \{0,\ 1,\ \ldots,\ n - 1\}\): \[\omega_k = \sqrt[n]{r}\left(\cos \frac{\varphi + 2\pi k}{n} + i\sin \frac{\varphi + 2\pi k}{n}\right).\]

1) \(z = 4\left(\cos \frac{16\pi}{19} + i\sin \frac{16\pi}{19}\right)\), \(n = 4\). Здесь \(\sqrt[4]{4} = \sqrt{2}\); угол \(\frac{\frac{16\pi}{19} + 2\pi k}{4} = \frac{4\pi}{19} + \frac{\pi k}{2}\), где \(k \in \{0,\ 1,\ 2,\ 3\}\): \[\begin{aligned} \omega_0 &= \sqrt{2}\left(\cos \frac{4\pi}{19} + i\sin \frac{4\pi}{19}\right), \\ \omega_1 &= \sqrt{2}\left(\cos \frac{27\pi}{38} + i\sin \frac{27\pi}{38}\right), \\ \omega_2 &= \sqrt{2}\left(\cos \frac{23\pi}{19} + i\sin \frac{23\pi}{19}\right), \\ \omega_3 &= \sqrt{2}\left(\cos \frac{65\pi}{38} + i\sin \frac{65\pi}{38}\right). \end{aligned}\] Здесь \(\frac{4\pi}{19} + \frac{\pi}{2} = \frac{27\pi}{38}\), \(\frac{4\pi}{19} + \pi = \frac{23\pi}{19}\), \(\frac{4\pi}{19} + \frac{3\pi}{2} = \frac{65\pi}{38}\).

2) \(z = 125\left(\cos \frac{9}{11} + i\sin \frac{9}{11}\right)\), \(n = 3\). Здесь \(\sqrt[3]{125} = 5\); угол \(\frac{\frac{9}{11} + 2\pi k}{3} = \frac{3}{11} + \frac{2\pi k}{3}\), где \(k \in \{0,\ 1,\ 2\}\): \[\begin{aligned} \omega_0 &= 5\left(\cos \frac{3}{11} + i\sin \frac{3}{11}\right), \\ \omega_1 &= 5\left(\cos\left(\frac{3}{11} + \frac{2\pi}{3}\right) + i\sin\left(\frac{3}{11} + \frac{2\pi}{3}\right)\right), \\ \omega_2 &= 5\left(\cos\left(\frac{3}{11} + \frac{4\pi}{3}\right) + i\sin\left(\frac{3}{11} + \frac{4\pi}{3}\right)\right). \end{aligned}\]

3) \(z = 64\), \(n = 6\). \(64 = 64(\cos 0 + i\sin 0)\), \(\sqrt[6]{64} = 2\); угол \(\frac{0 + 2\pi k}{6} = \frac{\pi k}{3}\), где \(k \in \{0,\ 1,\ 2,\ 3,\ 4,\ 5\}\): \[\begin{aligned} \omega_0 &= 2(\cos 0 + i\sin 0) = 2, \\ \omega_1 &= 2\left(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}\right) = 2\left(\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) = 1 + \sqrt{3}i, \\ \omega_2 &= 2\left(\cos \frac{2\pi}{3} + i\sin \frac{2\pi}{3}\right) = 2\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) = -1 + \sqrt{3}i, \\ \omega_3 &= 2(\cos \pi + i\sin \pi) = -2, \\ \omega_4 &= 2\left(\cos \frac{4\pi}{3} + i\sin \frac{4\pi}{3}\right) = 2\left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) = -1 - \sqrt{3}i, \\ \omega_5 &= 2\left(\cos \frac{5\pi}{3} + i\sin \frac{5\pi}{3}\right) = 2\left(\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) = 1 - \sqrt{3}i. \end{aligned}\]

4) \(z = \sqrt{3} - i\), \(n = 4\). \(|z| = \sqrt{3 + 1} = 2\), поэтому \[\sqrt{3} - i = 2\left(\frac{\sqrt{3}}{2} - \frac{1}{2}i\right) = 2\left(\cos\left(-\frac{\pi}{6}\right) + i\sin\left(-\frac{\pi}{6}\right)\right).\] Здесь \(\sqrt[4]{r} = \sqrt[4]{2}\); угол \(\frac{-\frac{\pi}{6} + 2\pi k}{4} = -\frac{\pi}{24} + \frac{\pi k}{2}\), где \(k \in \{0,\ 1,\ 2,\ 3\}\): \[\begin{aligned} \omega_0 &= \sqrt[4]{2}\left(\cos\left(-\frac{\pi}{24}\right) + i\sin\left(-\frac{\pi}{24}\right)\right), \\ \omega_1 &= \sqrt[4]{2}\left(\cos \frac{11\pi}{24} + i\sin \frac{11\pi}{24}\right), \\ \omega_2 &= \sqrt[4]{2}\left(\cos \frac{23\pi}{24} + i\sin \frac{23\pi}{24}\right), \\ \omega_3 &= \sqrt[4]{2}\left(\cos \frac{35\pi}{24} + i\sin \frac{35\pi}{24}\right). \end{aligned}\]