Упр.677 ГДЗ Колягин Ткачёва 11 класс (Алгебра)

1) корень 4 степени 4;
2) корень 3 степени i;
3) корень 6 степени 1;
4) корень 4 степени — 2 + 2i корень 3.

1. $$z^4=4=4(\cos 2\pi n+i\sin 2\pi n), \quad n\in \mathbb{Z}.$$
   Тогда
   $$z=\sqrt{2}\left(\cos \frac{\pi n}{2}+i\sin \frac{\pi n}{2}\right).$$
   При $n=0,1,2,3$ получаем:
   $$z_1=\sqrt{2}, \quad z_2=i\sqrt{2}, \quad z_3=-\sqrt{2}, \quad z_4=-i\sqrt{2}.$$
2. $$z^3=i=\cos\left(\frac{\pi}{2}+2\pi n\right)+i\sin\left(\frac{\pi}{2}+2\pi n\right), \quad n\in \mathbb{Z}.$$
   Тогда
   $$z=\cos\left(\frac{\pi}{6}+\frac{2\pi n}{3}\right)+i\sin\left(\frac{\pi}{6}+\frac{2\pi n}{3}\right).$$
   При $n=0,1,2$:
   $$z_1=\frac{\sqrt{3}}{2}+\frac{1}{2}i,$$
   $$z_2=-\frac{\sqrt{3}}{2}+\frac{1}{2}i,$$
   $$z_3=-i.$$
3. $$z^6=1=\cos 2\pi n+i\sin 2\pi n, \quad n\in \mathbb{Z}.$$
   Тогда
   $$z=\cos \frac{\pi n}{3}+i\sin \frac{\pi n}{3}.$$
   При $n=0,1,2,3,4,5$:
   $$z_1=1,$$
   $$z_2=\frac{1}{2}+\frac{\sqrt{3}}{2}i,$$
   $$z_3=-\frac{1}{2}+\frac{\sqrt{3}}{2}i,$$
   $$z_4=-1,$$
   $$z_5=-\frac{1}{2}-\frac{\sqrt{3}}{2}i,$$
   $$z_6=\frac{1}{2}-\frac{\sqrt{3}}{2}i.$$
4. $$-2+2i\sqrt{3}=4\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)=4\left(\cos \frac{2\pi}{3}+i\sin \frac{2\pi}{3}\right).$$
   Тогда
   $$z=\sqrt{2}\left(\cos\left(\frac{\pi}{6}+\frac{\pi n}{2}\right)+i\sin\left(\frac{\pi}{6}+\frac{\pi n}{2}\right)\right), \quad n\in \mathbb{Z}.$$
   При $n=0,1,2,3$:
   $$z_1=\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2}i,$$
   $$z_2=-\frac{\sqrt{2}}{2}+\frac{\sqrt{6}}{2}i,$$
   $$z_3=-\frac{\sqrt{6}}{2}-\frac{\sqrt{2}}{2}i,$$
   $$z_4=\frac{\sqrt{2}}{2}-\frac{\sqrt{6}}{2}i.$$

Ответ

1) $$\sqrt{2},\ i\sqrt{2},\ -\sqrt{2},\ -i\sqrt{2}$$

2) $$\frac{\sqrt{3}}{2}+\frac{1}{2}i,\ -\frac{\sqrt{3}}{2}+\frac{1}{2}i,\ -i$$

3) $$1,\ \frac{1}{2}+\frac{\sqrt{3}}{2}i,\ -\frac{1}{2}+\frac{\sqrt{3}}{2}i,\ -1,\ -\frac{1}{2}-\frac{\sqrt{3}}{2}i,\ \frac{1}{2}-\frac{\sqrt{3}}{2}i$$

4) $$\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2}i,\ -\frac{\sqrt{2}}{2}+\frac{\sqrt{6}}{2}i,\ -\frac{\sqrt{6}}{2}-\frac{\sqrt{2}}{2}i,\ \frac{\sqrt{2}}{2}-\frac{\sqrt{6}}{2}i$$