Menu Close

Question-165616




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

Leave a Reply

Your email address will not be published. Required fields are marked *