Question

Возвести в степень по формуле Муавра 1)(5√3+5i)^3
2) (-1-√3i)^6
3) (2+2i)^4
Расписать подробное решение

Answer 1

Ответ:

$1) (5\sqrt3+5i)^3 = 1000(cos(\frac{\pi}{2}) + isin(\frac{\pi}{2})) = 1000i\\\\2) (-1-i\sqrt3)^6 = 64(cos(0)+isin(0)) = 64\\\\3) (2+2i)^4 = 64(cos(\pi) + isin(\pi)) = -64$

Объяснение:

Формула Муавра:

$z^n = r^n(cos(n\phi)+isin(n\phi)), \ z = r(cos(\phi)+isin(\phi))$

$1.\\(5\sqrt3+5i)^3\\\\r = \sqrt{(5\sqrt3)^2+5^2} = \sqrt{75+25} = \sqrt{100} = 10\\\\\begin{cases}cos(\phi) = \frac{5\sqrt3}{10} = \frac{\sqrt3}{2}\\sin(\phi)=\frac{5}{10}=\frac{1}{2}\end{cases} \Rightarrow \phi = \frac{\pi}{6} \\\\z^3=(10(cos( \frac{\pi}{6})+isin( \frac{\pi}{6})))^3=10^3(cos( 3\cdot\frac{\pi}{6})+isin( 3\cdot\frac{\pi}{6}))=\\\\= 1000(cos( \frac{\pi}{2})+isint( \frac{\pi}{2})) = 1000(0+i) = 1000i\\\\2.\\(-1-i\sqrt3)^6\\\\r = \sqrt{(-1)^2+(-\sqrt3)^2}=\sqrt{1+3} = \sqrt4=2$

$\begin{cases} cos(\phi)=\frac{-1}{2}=-\frac{1}{2}\\sin(\phi)=\frac{-\sqrt3}{2}=-\frac{\sqrt3}{2} \end{cases}\Rightarrow \phi = \frac{4\pi}{3}$

$z^6 = 2^6(cos(6\cdot\frac{4\pi}{3})+isin(6\cdot\frac{4\pi}{3} )) = 64(cos(8\pi)+isin(8\pi))=\\\\ =64(cos(0)+isin(0)) = 64(1+0i) = 64\\\\3.\\(2+2i)^4\\\\r = \sqrt{2^2+2^2} = \sqrt{4+4} = \sqrt8 = 2\sqrt2\\\\\begin{equation*} \begin{cases} cos(\phi)=\frac{2}{2\sqrt2}=\frac{\sqrt2}{2} \\ sin(\phi)=\frac{2}{2\sqrt2}=\frac{\sqrt2}{2} \end{cases}\end{equation*} \Rightarrow\phi=\frac{\pi}{4} \\\\\\z^4=(2\sqrt2)^4(cos(4\cdot\frac{\pi}{4} )+isin(4\cdot\frac{\pi}{4})) = 64(cos(\pi)+isin(\pi)) =$

$= 64(-1+0i) = -64$