Упр.17.25 ГДЗ Никольский Потапов 11 класс (Алгебра)
- Найдите корни степени 3 из комплексного числа и найдите на комплексной плоскости точки, их изображающие: а) $$1$$; б) $$-1$$; в) $$8i$$; г) $$-8i$$; д) $$1+i$$; е) $$1-i$$; ж) $$1+\sqrt{3}$$; з) $$1-i\sqrt{3}$$.
$$z=1=\cos 0+i\sin 0$$
Тогда корни третьей степени:
$$\sqrt[3]{1}=\cos \frac{2\pi k}{3}+i\sin \frac{2\pi k}{3}, \quad k=0,1,2$$
$$\alpha_0=1,\quad \alpha_1=\cos \frac{2\pi}{3}+i\sin \frac{2\pi}{3}=-\frac12+\frac{\sqrt3}{2}i,\quad \alpha_2=\cos \frac{4\pi}{3}+i\sin \frac{4\pi}{3}=-\frac12-\frac{\sqrt3}{2}i$$
$$z=-1=\cos \pi+i\sin \pi$$
$$\sqrt[3]{-1}=\cos \frac{\pi+2\pi k}{3}+i\sin \frac{\pi+2\pi k}{3}, \quad k=0,1,2$$
$$\alpha_0=\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}=\frac12+\frac{\sqrt3}{2}i$$
$$\alpha_1=\cos \pi+i\sin \pi=-1$$
$$\alpha_2=\cos \frac{5\pi}{3}+i\sin \frac{5\pi}{3}=\frac12-\frac{\sqrt3}{2}i$$
$$z=8i=8\left(\cos \frac{\pi}{2}+i\sin \frac{\pi}{2}\right)$$
$$\sqrt[3]{8i}=2\left(\cos \frac{\frac{\pi}{2}+2\pi k}{3}+i\sin \frac{\frac{\pi}{2}+2\pi k}{3}\right), \quad k=0,1,2$$
$$\alpha_0=2\left(\cos \frac{\pi}{6}+i\sin \frac{\pi}{6}\right)=\sqrt3+i$$
$$\alpha_1=2\left(\cos \frac{5\pi}{6}+i\sin \frac{5\pi}{6}\right)=-\sqrt3+i$$
$$\alpha_2=2\left(\cos \frac{3\pi}{2}+i\sin \frac{3\pi}{2}\right)=-2i$$
$$z=-8i=8\left(\cos \frac{3\pi}{2}+i\sin \frac{3\pi}{2}\right)$$
$$\sqrt[3]{-8i}=2\left(\cos \frac{\frac{3\pi}{2}+2\pi k}{3}+i\sin \frac{\frac{3\pi}{2}+2\pi k}{3}\right), \quad k=0,1,2$$
$$\alpha_0=2\left(\cos \frac{\pi}{2}+i\sin \frac{\pi}{2}\right)=2i$$
$$\alpha_1=2\left(\cos \frac{7\pi}{6}+i\sin \frac{7\pi}{6}\right)=-\sqrt3-i$$
$$\alpha_2=2\left(\cos \frac{11\pi}{6}+i\sin \frac{11\pi}{6}\right)=\sqrt3-i$$
$$z=1+i=\sqrt2\left(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\right)$$
$$\sqrt[3]{1+i}=\sqrt[6]{2}\left(\cos \frac{\frac{\pi}{4}+2\pi k}{3}+i\sin \frac{\frac{\pi}{4}+2\pi k}{3}\right), \quad k=0,1,2$$
$$\alpha_0=\sqrt[6]{2}\left(\cos \frac{\pi}{12}+i\sin \frac{\pi}{12}\right)$$
$$\alpha_1=\sqrt[6]{2}\left(\cos \frac{3\pi}{4}+i\sin \frac{3\pi}{4}\right) =\sqrt[6]{2}\left(-\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}\right) =-\frac{1}{\sqrt[3]{2}}+\frac{i}{\sqrt[3]{2}}$$
$$\alpha_2=\sqrt[6]{2}\left(\cos \frac{17\pi}{12}+i\sin \frac{17\pi}{12}\right)$$
$$z=1-i=\sqrt2\left(\cos \frac{7\pi}{4}+i\sin \frac{7\pi}{4}\right)$$
$$\sqrt[3]{1-i}=\sqrt[6]{2}\left(\cos \frac{\frac{7\pi}{4}+2\pi k}{3}+i\sin \frac{\frac{7\pi}{4}+2\pi k}{3}\right), \quad k=0,1,2$$
$$\alpha_0=\sqrt[6]{2}\left(\cos \frac{7\pi}{12}+i\sin \frac{7\pi}{12}\right)$$
$$\alpha_1=\sqrt[6]{2}\left(\cos \frac{5\pi}{4}+i\sin \frac{5\pi}{4}\right) =\sqrt[6]{2}\left(-\frac{\sqrt2}{2}-i\frac{\sqrt2}{2}\right)$$
$$\alpha_2=\sqrt[6]{2}\left(\cos \frac{23\pi}{12}+i\sin \frac{23\pi}{12}\right)$$
$$z=1+i\sqrt3=2\left(\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}\right)$$
$$\sqrt[3]{1+i\sqrt3}=\sqrt[3]{2}\left(\cos \frac{\frac{\pi}{3}+2\pi k}{3}+i\sin \frac{\frac{\pi}{3}+2\pi k}{3}\right), \quad k=0,1,2$$
$$\alpha_0=\sqrt[3]{2}\left(\cos \frac{\pi}{9}+i\sin \frac{\pi}{9}\right)$$
$$\alpha_1=\sqrt[3]{2}\left(\cos \frac{7\pi}{9}+i\sin \frac{7\pi}{9}\right)$$
$$\alpha_2=\sqrt[3]{2}\left(\cos \frac{13\pi}{9}+i\sin \frac{13\pi}{9}\right)$$
$$z=1-i\sqrt3=2\left(\cos \frac{5\pi}{3}+i\sin \frac{5\pi}{3}\right)$$
$$\sqrt[3]{1-i\sqrt3}=\sqrt[3]{2}\left(\cos \frac{\frac{5\pi}{3}+2\pi k}{3}+i\sin \frac{\frac{5\pi}{3}+2\pi k}{3}\right), \quad k=0,1,2$$
$$\alpha_0=\sqrt[3]{2}\left(\cos \frac{5\pi}{9}+i\sin \frac{5\pi}{9}\right)$$
$$\alpha_1=\sqrt[3]{2}\left(\cos \frac{11\pi}{9}+i\sin \frac{11\pi}{9}\right)$$
$$\alpha_2=\sqrt[3]{2}\left(\cos \frac{17\pi}{9}+i\sin \frac{17\pi}{9}\right)$$
Ответ
а) $$1,\; -\frac12+\frac{\sqrt3}{2}i,\; -\frac12-\frac{\sqrt3}{2}i$$
б) $$\frac12+\frac{\sqrt3}{2}i,\; -1,\; \frac12-\frac{\sqrt3}{2}i$$
в) $$\sqrt3+i,\; -\sqrt3+i,\; -2i$$
г) $$2i,\; -\sqrt3-i,\; \sqrt3-i$$
д) $$\sqrt[6]{2}\left(\cos \frac{\pi}{12}+i\sin \frac{\pi}{12}\right),\; -\frac{1}{\sqrt[3]{2}}+\frac{i}{\sqrt[3]{2}},\; \sqrt[6]{2}\left(\cos \frac{17\pi}{12}+i\sin \frac{17\pi}{12}\right)$$
е) $$\sqrt[6]{2}\left(\cos \frac{7\pi}{12}+i\sin \frac{7\pi}{12}\right),\; \sqrt[6]{2}\left(\cos \frac{5\pi}{4}+i\sin \frac{5\pi}{4}\right),\; \sqrt[6]{2}\left(\cos \frac{23\pi}{12}+i\sin \frac{23\pi}{12}\right)$$
ж) $$\sqrt[3]{2}\left(\cos \frac{\pi}{9}+i\sin \frac{\pi}{9}\right),\; \sqrt[3]{2}\left(\cos \frac{7\pi}{9}+i\sin \frac{7\pi}{9}\right),\; \sqrt[3]{2}\left(\cos \frac{13\pi}{9}+i\sin \frac{13\pi}{9}\right)$$
з) $$\sqrt[3]{2}\left(\cos \frac{5\pi}{9}+i\sin \frac{5\pi}{9}\right),\; \sqrt[3]{2}\left(\cos \frac{11\pi}{9}+i\sin \frac{11\pi}{9}\right),\; \sqrt[3]{2}\left(\cos \frac{17\pi}{9}+i\sin \frac{17\pi}{9}\right)$$











