Задача-комплексни числа

[Гост]

Нужна ми е малко помощ

[Nathi123]

Нека [tex]z_{1 }=\frac{-2\sqrt{3}-6i}{6+2i\sqrt{3}}=\frac{-2\sqrt{3}(1+\sqrt{3}i)}{2\sqrt{3}(\sqrt{3}+i)}=\frac{-(1+\sqrt{3}i)(\sqrt{3}-i)}{(\sqrt{3}+i)(\sqrt{3}-i)}=-\frac{2\sqrt{3}+2i}{4}=-\frac{\sqrt{3}}{2}-\frac{1}{2}i = cos(-\frac{5\pi}{6})+isin(-\frac{5\pi}{6})[/tex]
[tex]\Rightarrow z=(\frac{-2\sqrt{3}-6i}{6+2i\sqrt{3}})^{879}=z_{1 }^{879}=cos(-293\frac{5\pi}{2})+isin(-293\frac{5\pi}{2})[/tex]
[tex]z= cos(-732\pi-\frac{\pi}{2})+isin(-732\pi-\frac{\pi}{2})=cos(-\frac{\pi}{2})+isin(-\frac{\pi}{2}) = 0 - i = - i[/tex]

[Nathi123]

2.зад. [tex]x^{241}+ 9\frac{\sqrt{3}}{2}+ 9i\frac{1}{2}=0 \Leftrightarrow x^{241} = 9( -\frac{\sqrt{3}}{2}-\frac{1}{2}i)[/tex]
Нека z = [tex]9( -\frac{\sqrt{3}}{2}-\frac{1}{2}i)=9( cos(-\frac{5\pi}{6})+isin(-\frac{5\pi}{6}))[/tex]
[tex]\Rightarrow x = \sqrt[241]{z} \Rightarrow x= \sqrt[241]{9} ( cos \frac{-\frac{5\pi}{6}+2k\pi}{241} + isin \frac{-\frac{5\pi}{6}+2k\pi}{241} ) ,
k = 0,1,2,...,240.[/tex]

[Гост]

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