Упр.654 ГДЗ Колягин Ткачёва 11 класс (Алгебра)

2) 8z3 — 27 = 0;
3) z4 = i;
4) z3 = -2i;
5) z8 = -2 + 2i;
6)z4-i=1.

1. $$z^4+81=0$$

   $$z^4=-81=81(\cos \pi+i\sin \pi)$$

   Тогда

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

   Получаем корни:

   $$z_1=\frac{3\sqrt2}{2}+\frac{3\sqrt2}{2}i,\quad z_2=-\frac{3\sqrt2}{2}+\frac{3\sqrt2}{2}i,\quad z_3=-\frac{3\sqrt2}{2}-\frac{3\sqrt2}{2}i,\quad z_4=\frac{3\sqrt2}{2}-\frac{3\sqrt2}{2}i.$$

2. $$8z^3-27=0$$

   $$z^3=\frac{27}{8}=\frac{27}{8}(\cos 2\pi+i\sin 2\pi)$$

   Тогда

   $$z=\frac32\left(\cos \frac{2\pi k}{3}+i\sin \frac{2\pi k}{3}\right),\quad k=0,1,2.$$

   Следовательно,

   $$z_1=\frac32,\quad z_2=-\frac34+\frac{3\sqrt3}{4}i,\quad z_3=-\frac34-\frac{3\sqrt3}{4}i.$$

3. $$z^4=i$$

   $$i=\cos \frac{\pi}{2}+i\sin \frac{\pi}{2}$$

   Тогда

   $$z=\cos \left(\frac{\pi}{8}+\frac{\pi k}{2}\right)+i\sin \left(\frac{\pi}{8}+\frac{\pi k}{2}\right),\quad k=0,1,2,3.$$

   Корни:

   $$z_1=\cos \frac{\pi}{8}+i\sin \frac{\pi}{8},\quad z_2=\cos \frac{5\pi}{8}+i\sin \frac{5\pi}{8},$$

   $$z_3=\cos \frac{9\pi}{8}+i\sin \frac{9\pi}{8},\quad z_4=\cos \frac{13\pi}{8}+i\sin \frac{13\pi}{8}.$$

4. $$z^3=-2i$$

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

   Тогда

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

   Следовательно,

   $$z_1=i\sqrt[3]{2},\quad z_2=\sqrt[3]{2}\left(-\frac{\sqrt3}{2}-\frac12 i\right),\quad z_3=\sqrt[3]{2}\left(\frac{\sqrt3}{2}-\frac12 i\right).$$

5. $$z^3=-2+2i$$

   $$-2+2i=2\sqrt2\left(\cos \frac{3\pi}{4}+i\sin \frac{3\pi}{4}\right)$$

   Тогда

   $$z=\sqrt[3]{2\sqrt2}\left(\cos \left(\frac{\pi}{4}+\frac{2\pi k}{3}\right)+i\sin \left(\frac{\pi}{4}+\frac{2\pi k}{3}\right)\right),\quad k=0,1,2.$$

   Так как $$\sqrt[3]{2\sqrt2}=\sqrt2,$$ получаем

   $$z_1=1+i,\quad z_2=\sqrt2\left(\cos \frac{11\pi}{12}+i\sin \frac{11\pi}{12}\right),\quad z_3=\sqrt2\left(\cos \frac{19\pi}{12}+i\sin \frac{19\pi}{12}\right).$$

6. $$z^4-i=1$$

   $$z^4=1+i=\sqrt2\left(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\right)$$

   Тогда

   $$z=\sqrt[8]{2}\left(\cos \left(\frac{\pi}{16}+\frac{\pi k}{2}\right)+i\sin \left(\frac{\pi}{16}+\frac{\pi k}{2}\right)\right),\quad k=0,1,2,3.$$

   Следовательно,

   $$z_1=\sqrt[8]{2}\left(\cos \frac{\pi}{16}+i\sin \frac{\pi}{16}\right),$$

   $$z_2=\sqrt[8]{2}\left(\cos \frac{9\pi}{16}+i\sin \frac{9\pi}{16}\right),$$

   $$z_3=\sqrt[8]{2}\left(\cos \frac{17\pi}{16}+i\sin \frac{17\pi}{16}\right),$$

   $$z_4=\sqrt[8]{2}\left(\cos \frac{25\pi}{16}+i\sin \frac{25\pi}{16}\right).$$

Ответ

1) $$z=\frac{3\sqrt2}{2}\pm \frac{3\sqrt2}{2}i,\; -\frac{3\sqrt2}{2}\pm \frac{3\sqrt2}{2}i$$

2) $$z=\frac32,\; -\frac34\pm \frac{3\sqrt3}{4}i$$

3) $$z=\cos \frac{\pi}{8}+i\sin \frac{\pi}{8},\; \cos \frac{5\pi}{8}+i\sin \frac{5\pi}{8},\; \cos \frac{9\pi}{8}+i\sin \frac{9\pi}{8},\; \cos \frac{13\pi}{8}+i\sin \frac{13\pi}{8}$$

4) $$z=i\sqrt[3]{2},\; \sqrt[3]{2}\left(-\frac{\sqrt3}{2}-\frac12 i\right),\; \sqrt[3]{2}\left(\frac{\sqrt3}{2}-\frac12 i\right)$$

5) $$z=1+i,\; \sqrt2\left(\cos \frac{11\pi}{12}+i\sin \frac{11\pi}{12}\right),\; \sqrt2\left(\cos \frac{19\pi}{12}+i\sin \frac{19\pi}{12}\right)$$

6) $$z=\sqrt[8]{2}\left(\cos \frac{\pi}{16}+i\sin \frac{\pi}{16}\right),\; \sqrt[8]{2}\left(\cos \frac{9\pi}{16}+i\sin \frac{9\pi}{16}\right),\; \sqrt[8]{2}\left(\cos \frac{17\pi}{16}+i\sin \frac{17\pi}{16}\right),\; \sqrt[8]{2}\left(\cos \frac{25\pi}{16}+i\sin \frac{25\pi}{16}\right)