Упр.17.15 ГДЗ Никольский Потапов 11 класс (Алгебра)

а) z= корень 3/2 +1/2*i;
б) z= корень 3/2 — 1/2*i;
в) z= -корень 3/2 +1/2*i;
г) z= -корень 3/2 — 1/2*i;
д) z= 1/2 + корень 3/2* i;
е) z= 1/2 — корень 3/2* i.

1. $$z=\frac{\sqrt{3}}{2}+\frac{1}{2}i=\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}.$$

   Тогда

   $$ \begin{aligned} z^1&=\frac{\sqrt{3}}{2}+\frac{1}{2}i,\\ z^2&=\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}=\frac12+\frac{\sqrt3}{2}i,\\ z^3&=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}=i,\\ z^4&=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}=-\frac12+\frac{\sqrt3}{2}i,\\ z^5&=\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}=-\frac{\sqrt3}{2}+\frac12 i,\\ z^6&=\cos\pi+i\sin\pi=-1,\\ z^7&=\cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6}=-\frac{\sqrt3}{2}-\frac12 i. \end{aligned} $$

2. $$z=\frac{\sqrt{3}}{2}-\frac{1}{2}i=\cos\frac{11\pi}{6}+i\sin\frac{11\pi}{6}.$$

   $$ \begin{aligned} z^1&=\frac{\sqrt{3}}{2}-\frac{1}{2}i,\\ z^2&=\cos\frac{11\pi}{3}+i\sin\frac{11\pi}{3}=\frac12-\frac{\sqrt3}{2}i,\\ z^3&=\cos\frac{11\pi}{2}+i\sin\frac{11\pi}{2}=-i,\\ z^4&=\cos\frac{22\pi}{3}+i\sin\frac{22\pi}{3}=-\frac12-\frac{\sqrt3}{2}i,\\ z^5&=\cos\frac{55\pi}{6}+i\sin\frac{55\pi}{6}=-\frac{\sqrt3}{2}+\frac12 i,\\ z^6&=\cos 11\pi+i\sin 11\pi=-1,\\ z^7&=\cos\frac{77\pi}{6}+i\sin\frac{77\pi}{6}=-\frac{\sqrt3}{2}+\frac12 i. \end{aligned} $$

3. $$z=-\frac{\sqrt{3}}{2}+\frac{1}{2}i=\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}.$$

   $$ \begin{aligned} z^1&=-\frac{\sqrt{3}}{2}+\frac{1}{2}i,\\ z^2&=\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3}=\frac12-\frac{\sqrt3}{2}i,\\ z^3&=\cos\frac{5\pi}{2}+i\sin\frac{5\pi}{2}=i,\\ z^4&=\cos\frac{10\pi}{3}+i\sin\frac{10\pi}{3}=-\frac12-\frac{\sqrt3}{2}i,\\ z^5&=\cos\frac{25\pi}{6}+i\sin\frac{25\pi}{6}=\frac{\sqrt3}{2}+\frac12 i,\\ z^6&=\cos 5\pi+i\sin 5\pi=-1,\\ z^7&=\cos\frac{35\pi}{6}+i\sin\frac{35\pi}{6}=\frac{\sqrt3}{2}+\frac12 i. \end{aligned} $$

4. $$z=-\frac{\sqrt{3}}{2}-\frac{1}{2}i=\cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6}.$$

   $$ \begin{aligned} z^1&=-\frac{\sqrt{3}}{2}-\frac{1}{2}i,\\ z^2&=\cos\frac{7\pi}{3}+i\sin\frac{7\pi}{3}=\frac12+\frac{\sqrt3}{2}i,\\ z^3&=\cos\frac{7\pi}{2}+i\sin\frac{7\pi}{2}=-i,\\ z^4&=\cos\frac{14\pi}{3}+i\sin\frac{14\pi}{3}=-\frac12+\frac{\sqrt3}{2}i,\\ z^5&=\cos\frac{35\pi}{6}+i\sin\frac{35\pi}{6}=\frac{\sqrt3}{2}-\frac12 i,\\ z^6&=\cos 7\pi+i\sin 7\pi=-1,\\ z^7&=\cos\frac{49\pi}{6}+i\sin\frac{49\pi}{6}=\frac{\sqrt3}{2}+\frac12 i. \end{aligned} $$

5. $$z=\frac12+\frac{\sqrt{3}}{2}i=\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}.$$

   $$ \begin{aligned} z^1&=\frac12+\frac{\sqrt{3}}{2}i,\\ z^2&=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}=-\frac12+\frac{\sqrt3}{2}i,\\ z^3&=\cos\pi+i\sin\pi=-1,\\ z^4&=\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}=-\frac12-\frac{\sqrt3}{2}i,\\ z^5&=\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3}=\frac12-\frac{\sqrt3}{2}i,\\ z^6&=\cos 2\pi+i\sin 2\pi=1,\\ z^7&=\cos\frac{7\pi}{3}+i\sin\frac{7\pi}{3}=\frac12+\frac{\sqrt3}{2}i. \end{aligned} $$

6. $$z=\frac12-\frac{\sqrt{3}}{2}i=\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3}.$$

   $$ \begin{aligned} z^1&=\frac12-\frac{\sqrt{3}}{2}i,\\ z^2&=\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}=-\frac12-\frac{\sqrt3}{2}i,\\ z^3&=\cos 5\pi+i\sin 5\pi=-1,\\ z^4&=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}=-\frac12+\frac{\sqrt3}{2}i,\\ z^5&=\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}=\frac12+\frac{\sqrt3}{2}i,\\ z^6&=\cos 10\pi+i\sin 10\pi=1,\\ z^7&=\cos\frac{35\pi}{3}+i\sin\frac{35\pi}{3}=\frac12-\frac{\sqrt3}{2}i. \end{aligned} $$

Ответ

а) $$\frac{\sqrt3}{2}+\frac12 i,\ \frac12+\frac{\sqrt3}{2}i,\ i,\ -\frac12+\frac{\sqrt3}{2}i,\ -\frac{\sqrt3}{2}+\frac12 i,\ -1,\ -\frac{\sqrt3}{2}-\frac12 i.$$
б) $$\frac{\sqrt3}{2}-\frac12 i,\ \frac12-\frac{\sqrt3}{2}i,\ -i,\ -\frac12-\frac{\sqrt3}{2}i,\ -\frac{\sqrt3}{2}+\frac12 i,\ -1,\ -\frac{\sqrt3}{2}+\frac12 i.$$
в) $$-\frac{\sqrt3}{2}+\frac12 i,\ \frac12-\frac{\sqrt3}{2}i,\ i,\ -\frac12-\frac{\sqrt3}{2}i,\ \frac{\sqrt3}{2}+\frac12 i,\ -1,\ \frac{\sqrt3}{2}+\frac12 i.$$
г) $$-\frac{\sqrt3}{2}-\frac12 i,\ \frac12+\frac{\sqrt3}{2}i,\ -i,\ -\frac12+\frac{\sqrt3}{2}i,\ \frac{\sqrt3}{2}-\frac12 i,\ -1,\ \frac{\sqrt3}{2}+\frac12 i.$$
д) $$\frac12+\frac{\sqrt3}{2}i,\ -\frac12+\frac{\sqrt3}{2}i,\ -1,\ -\frac12-\frac{\sqrt3}{2}i,\ \frac12-\frac{\sqrt3}{2}i,\ 1,\ \frac12+\frac{\sqrt3}{2}i.$$
е) $$\frac12-\frac{\sqrt3}{2}i,\ -\frac12-\frac{\sqrt3}{2}i,\ -1,\ -\frac12+\frac{\sqrt3}{2}i,\ \frac12+\frac{\sqrt3}{2}i,\ 1,\ \frac12-\frac{\sqrt3}{2}i.$$