Задача 72018 (-1+isqrt(3))^9 z =...
Условие
z = 2(cos2Pi+isin2Pi)
(3/2 - sqrt(3)/2 i)^3
Решение
[m](-1+i\sqrt{3})^{9} = [2(-\frac{1}{2} + i\frac{\sqrt{3}}{2})]^{9} =[2(cos \frac{2\pi}{3} + i \cdot sin\frac{2\pi}{3})]^{9} = [/m]
[m] =2^{9}(cos\frac{18\pi}{3} + i \cdot sin\frac{18\pi}{3}) = 2^{9}(cos(6\pi) + i \cdot sin(6\pi)) =2^{9}(1+i \cdot 0) = 512[/m]
8) z = 2(cos 2π + i*sin 2π) = 2*(cos 0 + i*sin 0) = 2*(1 + 0) = 2
9) Тоже по формуле Муавра.
[m](\frac{3}{2} - \frac{\sqrt{3}}{2} \cdot i)^3 = [\sqrt{3}(\frac{\sqrt{3}}{2} - \frac{1}{2}\cdot i)]^3 = [\sqrt{3}(cos \frac{11\pi}{6} + i \cdot sin \frac{11\pi}{6})]^3 =[/m]
[m]= (\sqrt{3})^3(cos \frac{11\pi}{2} + i \cdot sin \frac{11\pi}{2}) = \sqrt{27}(cos \frac{3\pi}{2} + i \cdot sin \frac{3\pi}{2}) = \sqrt{27}(0 + i(-1)) = -\sqrt{27} \cdot i[/m]