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Question Number 165616 by amin96 last updated on 05/Feb/22

Commented by mkam last updated on 06/Feb/22

find all the root of ( x^9 − 1 = 0) ? ⊂ solution ⊃ x^9 = 1 ⇒ x = (1)^(1/9) ⇒ x_k = r^n ( cos ((𝛉 +2k𝛑)/n) + i sin ((𝛉 + 2k𝛑)/n) ) put : r = (√(x^2 +y^2 )) = (√((1)^2 +(0)^2 )) = (√( 1)) = 1 𝛉 = tan^(−1) ((y/x)) = tan^(−1) ((0/1) ) = tan^(−1) (0) = 0 n = 9 , k = 0,1,2,........,8 k = 0 x_0 = cos(0) + i sin(0) = 1 k = 1 x_1 = cos ((2𝛑)/9) + i sin ((2𝛑)/9) k = 2 x_2 = cos ((4𝛑)/9) + i sin ((4𝛑)/9) k = 3 x_3 = cos ((2𝛑)/3) + i sin ((2𝛑)/3) = − (1/2) + ((√3)/2)i k = 4 x_4 = cos ((8𝛑)/9) + i sin ((8𝛑)/9) k = 5 x_5 = cos ((10𝛑)/9) + i sin ((10𝛑)/9) k = 6 x_6 = cos ((4𝛑)/3) + i sin ((4𝛑)/3) = − (1/2) − ((√3)/2) i k = 7 x_7 = cos ((14𝛑)/9) + i sin ((14𝛑)/9) k = 8 x_8 = cos ((16𝛑)/9) + i sin ((16𝛑)/9) (mohammad aldolimy)

$$\boldsymbol{{find}}\:\boldsymbol{{all}}\:\boldsymbol{{the}}\:\boldsymbol{{root}}\:\boldsymbol{{of}}\:\left(\:\boldsymbol{{x}}^{\mathrm{9}} \:−\:\mathrm{1}\:=\:\mathrm{0}\right)\:? \\ $$$$ \\ $$$$\subset\:\boldsymbol{{solution}}\:\supset \\ $$$$ \\ $$$$\boldsymbol{{x}}^{\mathrm{9}} \:=\:\mathrm{1}\:\Rightarrow\:\boldsymbol{{x}}\:=\:\left(\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{9}}} \:\Rightarrow\:\boldsymbol{{x}}_{\boldsymbol{{k}}} \:=\:\boldsymbol{{r}}^{\boldsymbol{{n}}} \left(\:\boldsymbol{{cos}}\:\frac{\boldsymbol{\theta}\:+\mathrm{2}\boldsymbol{{k}\pi}}{\boldsymbol{{n}}}\:+\:\boldsymbol{{i}}\:\boldsymbol{{sin}}\:\frac{\boldsymbol{\theta}\:+\:\mathrm{2}\boldsymbol{{k}\pi}}{\boldsymbol{{n}}}\:\right) \\ $$$$ \\ $$$$\boldsymbol{{put}}\::\:\boldsymbol{{r}}\:=\:\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} }\:=\:\sqrt{\left(\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{0}\right)^{\mathrm{2}} }\:=\:\sqrt{\:\mathrm{1}}\:=\:\mathrm{1} \\ $$$$ \\ $$$$\boldsymbol{\theta}\:=\:\boldsymbol{{tan}}^{−\mathrm{1}} \left(\frac{\boldsymbol{{y}}}{\boldsymbol{{x}}}\right)\:=\:\boldsymbol{{tan}}^{−\mathrm{1}} \left(\frac{\mathrm{0}}{\mathrm{1}}\:\right)\:=\:\boldsymbol{{tan}}^{−\mathrm{1}} \left(\mathrm{0}\right)\:=\:\mathrm{0} \\ $$$$ \\ $$$$\boldsymbol{{n}}\:=\:\mathrm{9}\:,\:\boldsymbol{{k}}\:=\:\mathrm{0},\mathrm{1},\mathrm{2},........,\mathrm{8} \\ $$$$ \\ $$$$\boldsymbol{{k}}\:=\:\mathrm{0} \\ $$$$\boldsymbol{{x}}_{\mathrm{0}} \:=\:\boldsymbol{{cos}}\left(\mathrm{0}\right)\:+\:\boldsymbol{{i}}\:\boldsymbol{{sin}}\left(\mathrm{0}\right)\:=\:\mathrm{1} \\ $$$$\boldsymbol{{k}}\:=\:\mathrm{1} \\ $$$$\boldsymbol{{x}}_{\mathrm{1}} \:=\:\boldsymbol{{cos}}\:\frac{\mathrm{2}\boldsymbol{\pi}}{\mathrm{9}}\:+\:\boldsymbol{{i}}\:\boldsymbol{{sin}}\:\frac{\mathrm{2}\boldsymbol{\pi}}{\mathrm{9}} \\ $$$$\boldsymbol{{k}}\:=\:\mathrm{2} \\ $$$$\boldsymbol{{x}}_{\mathrm{2}} \:=\:\boldsymbol{{cos}}\:\frac{\mathrm{4}\boldsymbol{\pi}}{\mathrm{9}}\:+\:\boldsymbol{{i}}\:\boldsymbol{{sin}}\:\frac{\mathrm{4}\boldsymbol{\pi}}{\mathrm{9}} \\ $$$$\boldsymbol{{k}}\:=\:\mathrm{3} \\ $$$$\boldsymbol{{x}}_{\mathrm{3}} \:=\:\boldsymbol{{cos}}\:\frac{\mathrm{2}\boldsymbol{\pi}}{\mathrm{3}}\:+\:\boldsymbol{{i}}\:\boldsymbol{{sin}}\:\frac{\mathrm{2}\boldsymbol{\pi}}{\mathrm{3}}\:=\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\boldsymbol{{i}} \\ $$$$\boldsymbol{{k}}\:=\:\mathrm{4} \\ $$$$\boldsymbol{{x}}_{\mathrm{4}} \:=\:\boldsymbol{{cos}}\:\frac{\mathrm{8}\boldsymbol{\pi}}{\mathrm{9}}\:+\:\boldsymbol{{i}}\:\boldsymbol{{sin}}\:\frac{\mathrm{8}\boldsymbol{\pi}}{\mathrm{9}} \\ $$$$\boldsymbol{{k}}\:=\:\mathrm{5} \\ $$$$\boldsymbol{{x}}_{\mathrm{5}} \:=\:\boldsymbol{{cos}}\:\frac{\mathrm{10}\boldsymbol{\pi}}{\mathrm{9}}\:+\:\boldsymbol{{i}}\:\boldsymbol{{sin}}\:\frac{\mathrm{10}\boldsymbol{\pi}}{\mathrm{9}} \\ $$$$\boldsymbol{{k}}\:=\:\mathrm{6} \\ $$$$\boldsymbol{{x}}_{\mathrm{6}} \:=\:\boldsymbol{{cos}}\:\frac{\mathrm{4}\boldsymbol{\pi}}{\mathrm{3}}\:+\:\boldsymbol{{i}}\:\boldsymbol{{sin}}\:\frac{\mathrm{4}\boldsymbol{\pi}}{\mathrm{3}}\:=\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:−\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\boldsymbol{{i}} \\ $$$$\boldsymbol{{k}}\:=\:\mathrm{7} \\ $$$$\boldsymbol{{x}}_{\mathrm{7}} \:=\:\boldsymbol{{cos}}\:\frac{\mathrm{14}\boldsymbol{\pi}}{\mathrm{9}}\:+\:\boldsymbol{{i}}\:\boldsymbol{{sin}}\:\frac{\mathrm{14}\boldsymbol{\pi}}{\mathrm{9}} \\ $$$$\boldsymbol{{k}}\:=\:\mathrm{8} \\ $$$$\boldsymbol{{x}}_{\mathrm{8}} \:=\:\boldsymbol{{cos}}\:\frac{\mathrm{16}\boldsymbol{\pi}}{\mathrm{9}}\:+\:\boldsymbol{{i}}\:\boldsymbol{{sin}}\:\frac{\mathrm{16}\boldsymbol{\pi}}{\mathrm{9}} \\ $$$$ \\ $$$$\left(\boldsymbol{{mohammad}}\:\boldsymbol{{aldolimy}}\right) \\ $$

Answered by Ar Brandon last updated on 05/Feb/22

x^9 −1=0 ⇒x^9 =e^(2πik) ⇒x=e^(((2k)/9)iπ) , k∈[0, 8]

$${x}^{\mathrm{9}} −\mathrm{1}=\mathrm{0}\:\Rightarrow{x}^{\mathrm{9}} ={e}^{\mathrm{2}\pi{ik}} \:\Rightarrow{x}={e}^{\frac{\mathrm{2}{k}}{\mathrm{9}}{i}\pi} \:,\:{k}\in\left[\mathrm{0},\:\mathrm{8}\right] \\ $$

Answered by alephzero last updated on 05/Feb/22

x^9 −1 = 0 x^9 = 1 x = 1

$${x}^{\mathrm{9}} −\mathrm{1}\:=\:\mathrm{0} \\ $$$${x}^{\mathrm{9}} \:=\:\mathrm{1} \\ $$$${x}\:=\:\mathrm{1} \\ $$

Commented by mkam last updated on 06/Feb/22

false

$${false} \\ $$

Commented by alephzero last updated on 11/Feb/22

Why? This is just one of all solutions.

$${Why}?\:{This}\:{is}\:{just}\:\boldsymbol{{one}}\:{of}\:{all} \\ $$$${solutions}. \\ $$

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