Решить уравнения на множестве комплексных чисел. x^6+1=0

x^5-1=0

x^3+8=0

4x^4+1=0
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Ответ:

alexey1009

06.01.2021 10:56

$x^6+1=0\\\\x^6=-1\\\\x=\sqrt[6]{-1}\\\\x=(cos(\pi)+isin(\pi))^{\frac{1}{6} }\\\\x =cos(\frac{\pi}{6}+\frac{2\pi k}{6})+isin(\frac{\pi}{6}+\frac{2\pi k}{6}),\ k\in\mathbb Z\\\\x=cos(\frac{\pi}{6}+\frac{\pi k}{3})+isin(\frac{\pi}{6}+\frac{\pi k}{3} ), \ k \in \mathbb Z\\\\x_1 = cos(\frac{\pi}{6} ) + isin(\frac{\pi}{6}) = \frac{\sqrt3}{2} +\frac{1}{2}i\\\\x_2 = cos(\frac{\pi}{6}+\frac{\pi}{3} ) + isin(\frac{\pi}{6}+\frac{\pi}{3} ) = cos(\frac{\pi}{2} ) + isin(\frac{\pi}{2})=0+i=i$

$x_3=cos(\frac{\pi}{6}+\frac{2\pi}{3})+isin(\frac{\pi}{6}+\frac{2\pi}{3})=cos(\frac{5\pi}{6})+isin(\frac{5\pi}{6})=-\frac{\sqrt3}{2}+\frac{1}{2}i\\\\x_4=cos(\frac{\pi}{6}+\frac{3\pi}{3})+isin(\frac{\pi}{6}+\frac{3\pi}{3})=cos(\frac{7\pi}{6})+isin(\frac{7\pi}{6})=-\frac{\sqrt3}{2}-\frac{1}{2}i\\\\x_5=cos(\frac{\pi}{6}+\frac{4\pi}{3})+isin(\frac{\pi}{6}+\frac{4\pi}{3})=cos(\frac{3\pi}{2})+isin(\frac{3\pi}{2})=0-i = -i$

$x_6=cos(\frac{\pi}{6}+\frac{5\pi}{3})+isin(\frac{\pi}{6}+\frac{5\pi}{3})=cos(\frac{11\pi}{6})+isin(\frac{11\pi}{6})=\frac{\sqrt3}{2}-\frac{1}{2}i$

$x^5-1=0\\\\x^5=1\\\\x=\sqrt[5]{1}\\\\x=(cos(0)+isin(0))^{\frac{1}{5}}\\\\x=cos(\frac{2\pi k}{5})+isin(\frac{2\pi k}{5}),\,k\in\mathbb Z\\\\x_1=cos(0)+isin(0)=1+0i=1\\\\x_2=cos(\frac{2\pi}{5})+isin(\frac{2\pi}{5})\\\\x_3=cos(\frac{4\pi}{5})+isin(\frac{4\pi}{5})\\\\x_4=cos(\frac{6\pi}{5})+isin(\frac{6\pi}{5})\\\\x_5=cos(\frac{8\pi}{5})+isin(\frac{8\pi}{5})$

$x^3+8=0\\\\x^3=-8\\\\x=\sqrt[3]{-8}\\\\x=(8(cos(\pi)+isin(\pi)))^{\frac{1}{3}}\\\\x=\sqrt[3]{8}(cos(\frac{\pi}{3}+\frac{2\pi k}{3})+isin(\frac{\pi}{3}+\frac{2\pi k}{3})),\,k\in\mathbb Z\\\\x=2(cos(\frac{\pi}{3}+\frac{2\pi k}{3})+isin(\frac{\pi}{3}+\frac{2\pi k}{3})),\,k\in\mathbb Z\\\\x_1=2(cos(\frac{\pi}{3})+isin(\frac{\pi}{3}))=1+i\sqrt3\\\\x_2=2(cos(\frac{\pi}{3}+\frac{2\pi}{3})+isin(\frac{\pi}{3}+\frac{2\pi}{3}))=2(cos(\pi)+isin(\pi)) = -2+0i=-2$

$x_3=2(cos(\frac{\pi}{3}+\frac{4\pi}{3})+isin(\frac{\pi}{3}+\frac{4\pi}{3}))=2(cos(\frac{5\pi}{3}+isin(\frac{5\pi}{3})) = 1-\sqrt3i$

$4x^4+1=0\\\\4x^4=-1\\\\x^4=-\frac{1}{4}\\\\x=\sqrt[4]{-\frac{1}{4}}\\\\x=(\frac{1}{4}(cos(\pi)+isin(\pi)))^{\frac{1}{4}}\\\\x=\frac{\sqrt2}{2}(cos(\frac{\pi}{4}+\frac{2\pi k}{4})+isin(\frac{\pi}{4}+\frac{2\pi k}{4})),\,k\in\mathbb Z\\\\x=\frac{\sqrt2}{2}(cos(\frac{\pi}{4}+\frac{\pi k}{2})+isin(\frac{\pi}{4}+\frac{\pi k}{2})),\,k\in\mathbb Z\\\\x_1=\frac{\sqrt2}{2}(cos(\frac{\pi}{4})+isin(\frac{\pi}{4}))=\frac{1}{2}+\frac{1}{2}i\\\\$

$x_2=\frac{\sqrt2}{2}(cos(\frac{\pi}{4}+\frac{\pi}{2})+isin(\frac{\pi}{4}+\frac{\pi}{2}))=\frac{\sqrt2}{2}(cos(\frac{3\pi}{4})+isin(\frac{3\pi}{4}))=-\frac{1}{2}+\frac{1}{2}i\\\\x_3=\frac{\sqrt2}{2}(cos(\frac{\pi}{4}+\frac{2\pi}{2})+isin(\frac{\pi}{4}+\frac{2\pi}{2}))=\frac{\sqrt2}{2}(cos(\frac{5\pi}{4})+isin(\frac{5\pi}{4}))=-\frac{1}{2}-\frac{1}{2}i$

$x_4=\frac{\sqrt2}{2}(cos(\frac{\pi}{4}+\frac{3\pi}{2})+isin(\frac{\pi}{4}+\frac{3\pi}{2}))=\frac{\sqrt2}{2}(cos(\frac{7\pi}{4})+isin(\frac{7\pi}{4}))=\frac{1}{2}-\frac{1}{2}i$