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Question Number 94479 by i jagooll last updated on 19/May/20

$$\mathrm{find}\mathrm{solution}\mathrm{in}\mathbb{C}$$$${\mathrm{x}}^4+{\mathrm{x}}^3+{\mathrm{x}}^2+\mathrm{x}+1=0$$

Answered by john santu last updated on 19/May/20

$$\Rightarrow (x-1)(x^4+x^3+x^2+x+1)=0$$$$\Rightarrow x^5-1=0$$$$x^5=1\Rightarrow {\left(x^5\right)}^{\frac{1}{5}}={\left(e^{2\pi i}\right)}^{\frac{1}{5}}$$$$\Rightarrow x=e^{\left(\frac{2\pi +2\pi \mathrm{n}}{5}\right).i}$$$$x_1=\cos \left(\frac{2\pi }{5}\right)+i\sin \left(\frac{2\pi }{5}\right)$$$$x_2=\cos \left(\frac{4\pi }{5}\right)+i\sin \left(\frac{4\pi }{5}\right)$$$$x_3=\cos \left(\frac{6\pi }{5}\right)+i\sin \left(\frac{6\pi }{5}\right)$$$$x_4=\cos \left(\frac{8\pi }{5}\right)+i\sin \left(\frac{8\pi }{5}\right)$$

Commented by i jagooll last updated on 19/May/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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