Упр.15.10 ГДЗ Мерзляк 11 класс Углубленный уровень (Алгебра)

1) z=4(cos(16п/19)+isin(16п/19)), n=4; 3) z=64, n=6;
2) z=125(cos(9/11)+isin(9/11)), n=3; 4) z=v3-i, n=4.

1. $$z=4\left(\cos \frac{16\pi}{19}+i\sin \frac{16\pi}{19}\right), \quad n=4.$$

   По формуле корней $$n$$-й степени:

   $$
   \sqrt[4]{z}= \sqrt[4]{4}\left(\cos \frac{\frac{16\pi}{19}+2\pi k}{4}+i\sin \frac{\frac{16\pi}{19}+2\pi k}{4}\right), \quad k=0,1,2,3.
   $$

   Так как $$\sqrt[4]{4}=\sqrt{2},$$ получаем:

   $$
   \sqrt[4]{z_k}=\sqrt{2}\left(\cos \left(\frac{4\pi}{19}+\frac{\pi k}{2}\right)+i\sin \left(\frac{4\pi}{19}+\frac{\pi k}{2}\right)\right), \quad k=0,1,2,3.
   $$

   Следовательно, корни:

   $$\sqrt{2}\left(\cos \frac{4\pi}{19}+i\sin \frac{4\pi}{19}\right),$$
   $$\sqrt{2}\left(\cos \frac{27\pi}{38}+i\sin \frac{27\pi}{38}\right),$$
   $$\sqrt{2}\left(\cos \frac{23\pi}{19}+i\sin \frac{23\pi}{19}\right),$$
   $$\sqrt{2}\left(\cos \frac{65\pi}{38}+i\sin \frac{65\pi}{38}\right).$$

2. $$z=125\left(\cos \frac{9}{11}+i\sin \frac{9}{11}\right), \quad n=3.$$

   По формуле корней $$3$$-й степени:

   $$
   \sqrt[3]{z}= \sqrt[3]{125}\left(\cos \frac{\frac{9}{11}+2\pi k}{3}+i\sin \frac{\frac{9}{11}+2\pi k}{3}\right), \quad k=0,1,2.
   $$

   Так как $$\sqrt[3]{125}=5,$$ получаем:

   $$
   \sqrt[3]{z_k}=5\left(\cos \left(\frac{3}{11}+\frac{2\pi k}{3}\right)+i\sin \left(\frac{3}{11}+\frac{2\pi k}{3}\right)\right), \quad k=0,1,2.
   $$

   Следовательно, корни:

   $$5\left(\cos \frac{3}{11}+i\sin \frac{3}{11}\right),$$
   $$5\left(\cos \left(\frac{3}{11}+\frac{2\pi}{3}\right)+i\sin \left(\frac{3}{11}+\frac{2\pi}{3}\right)\right),$$
   $$5\left(\cos \left(\frac{3}{11}+\frac{4\pi}{3}\right)+i\sin \left(\frac{3}{11}+\frac{4\pi}{3}\right)\right).$$

3. $$z=64, \quad n=6.$$

   Представим число в тригонометрической форме:

   $$64=64(\cos 0+i\sin 0).$$

   Тогда

   $$
   \sqrt[6]{z}= \sqrt[6]{64}\left(\cos \frac{0+2\pi k}{6}+i\sin \frac{0+2\pi k}{6}\right), \quad k=0,1,2,3,4,5.
   $$

   Так как $$\sqrt[6]{64}=2,$$ имеем:

   $$
   \sqrt[6]{z_k}=2\left(\cos \frac{\pi k}{3}+i\sin \frac{\pi k}{3}\right), \quad k=0,1,2,3,4,5.
   $$

   Корни:

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

4. $$z=\sqrt{3}-i, \quad n=4.$$

   Найдём модуль и аргумент числа:

   $$
   r=\sqrt{(\sqrt{3})^2+(-1)^2}=\sqrt{4}=2,
   $$

   $$
   \cos \varphi=\frac{\sqrt{3}}{2}, \quad \sin \varphi=-\frac{1}{2},
   $$

   значит,

   $$z=2\left(\cos \left(-\frac{\pi}{6}\right)+i\sin \left(-\frac{\pi}{6}\right)\right).$$

   Тогда

   $$
   \sqrt[4]{z}= \sqrt[4]{2}\left(\cos \left(\frac{-\pi/6+2\pi k}{4}\right)+i\sin \left(\frac{-\pi/6+2\pi k}{4}\right)\right), \quad k=0,1,2,3.
   $$

   То есть

   $$
   \sqrt[4]{z_k}= \sqrt[4]{2}\left(\cos \left(-\frac{\pi}{24}+\frac{\pi k}{2}\right)+i\sin \left(-\frac{\pi}{24}+\frac{\pi k}{2}\right)\right), \quad k=0,1,2,3.
   $$

   Следовательно, корни:

   $$\sqrt[4]{2}\left(\cos \left(-\frac{\pi}{24}\right)+i\sin \left(-\frac{\pi}{24}\right)\right),$$
   $$\sqrt[4]{2}\left(\cos \frac{11\pi}{24}+i\sin \frac{11\pi}{24}\right),$$
   $$\sqrt[4]{2}\left(\cos \frac{23\pi}{24}+i\sin \frac{23\pi}{24}\right),$$
   $$\sqrt[4]{2}\left(\cos \frac{35\pi}{24}+i\sin \frac{35\pi}{24}\right).$$

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) $$2,\ 1+i\sqrt{3},\ -1+i\sqrt{3},\ -2,\ -1-i\sqrt{3},\ 1-i\sqrt{3};$$
4) $$\sqrt[4]{2}\left(\cos \left(-\frac{\pi}{24}\right)+i\sin \left(-\frac{\pi}{24}\right)\right),\ \sqrt[4]{2}\left(\cos \frac{11\pi}{24}+i\sin \frac{11\pi}{24}\right),\ \sqrt[4]{2}\left(\cos \frac{23\pi}{24}+i\sin \frac{23\pi}{24}\right),\ \sqrt[4]{2}\left(\cos \frac{35\pi}{24}+i\sin \frac{35\pi}{24}\right).$$