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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)

findalltherootof(x91=0)?solutionx9=1x=(1)19xk=rn(cosθ+2kπn+isinθ+2kπn)put:r=x2+y2=(1)2+(0)2=1=1θ=tan1(yx)=tan1(01)=tan1(0)=0n=9,k=0,1,2,........,8k=0x0=cos(0)+isin(0)=1k=1x1=cos2π9+isin2π9k=2x2=cos4π9+isin4π9k=3x3=cos2π3+isin2π3=12+32ik=4x4=cos8π9+isin8π9k=5x5=cos10π9+isin10π9k=6x6=cos4π3+isin4π3=1232ik=7x7=cos14π9+isin14π9k=8x8=cos16π9+isin16π9(mohammadaldolimy)

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]

x91=0x9=e2πikx=e2k9iπ,k[0,8]

Answered by alephzero last updated on 05/Feb/22

x^9 −1 = 0  x^9  = 1  x = 1

x91=0x9=1x=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?Thisisjustoneofallsolutions.

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