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Class 12 Tuition > Prove that sum of cube roots of unity is zero.
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Prove that sum of cube roots of unity is zero.

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We are mainly interested in the roots of unity; however, we begin by being general and presenting a theorem. The cyclotomic equation is: cyclotomic Where , n=2,3... and in the complex plane. And because it is not surprisingly, cyclotomic means 'related to the roots of unity' and often refers...
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We are mainly interested in the roots of unity; however, we begin by being general and presenting a theorem. The cyclotomic equation is: cyclotomic[1.01] Where [1.02], n=2,3... and in the complex plane. And because it is not surprisingly, cyclotomic means 'related to the roots of unity' and often refers to the equation: [1.03] The theorem is that the roots of cyclotomicare: [1.04] Where z and a are complex numbers, l=0,1,2, ... ,n-1, and ??read less
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Maths Teacher

Cube roots of unity are 1, (-1+root 3i)/2 and (-1-root3i)/2. If you add all these you get zero.If you factorise x^3-1 yiy get (x-1) (x^2+x+1) . And the roots of tge second equation are complex as given above
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Mathematics

Consider an equation: z3-1=0 here three roots are : z1,z2,z3 then z3-1=0 z3=1 then we can easily solve z1=1, z2=w= (-1+i?3)/2, & z3=w2= (-1-i?3)/2 HERE WE CAN EASILY SOLVE THAT 1+Z2+Z3 =1+W+W2=1+ (-1+i?3)/2 + (-1-i?3)/2 =(2-1+i?3)/2 (-1-i?3)/...
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Consider an equation: z3-1=0 here three roots are : z1,z2,z3 then z3-1=0 z3=1 then we can easily solve z1=1, z2=w= (-1+i?3)/2, & z3=w2= (-1-i?3)/2 HERE WE CAN EASILY SOLVE THAT 1+Z2+Z3 =1+W+W2=1+ (-1+i?3)/2 + (-1-i?3)/2 =(2-1+i?3)/2 (-1-i?3)/ = 0 i.e. 1+w+w2=0 SO, sum of cube roots of unity is zero read less
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Physics and Maths made easy

Write 1.0 = 1.0* Therefore (1.0)^(1/3) = 1.0 * for n={0,1,2} Therefore the cube roots are 1.0, (1/2) + i*sq root (3))/2, (-1/2)+ i*(- sq root (3))/2] sum of which three roots is + i * = 0+i*0
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Write 1.0 = 1.0*[cos(2*n*pi) + sin(2*n*pi)] Therefore (1.0)^(1/3) = 1.0 * [cos(2*n*pi/3)+sin(2*n*pi/3)] for n={0,1,2} Therefore the cube roots are 1.0, (1/2) + i*sq root (3))/2, (-1/2)+ i*(- sq root (3))/2] sum of which three roots is [1+(1/2)+(-1/2)] + i * [(sq root (3))/2 - (sq root (3))/2] = 0+i*0 read less
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1+w+w2=0
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SME in Mathematics & Statistics

To find roots of the equation x3 - 1 = 0 x3 - 1 = 0 (x - 1)(x2 + x + 1) = 0 Since x3 - 1 = 0 is a polynomial equation of degree 3, it will have three roots. We can see that one of the roots is 1. The other 2 roots are complex roots. Let one of them is w. Then it will satisfy the equation. (w -...
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To find roots of the equation x3 - 1 = 0 x3 - 1 = 0 (x - 1)(x2 + x + 1) = 0 Since x3 - 1 = 0 is a polynomial equation of degree 3, it will have three roots. We can see that one of the roots is 1. The other 2 roots are complex roots. Let one of them is w. Then it will satisfy the equation. (w - 1)(w2 + w + 1) = 0 w cannot be 1. Hence, w2 + w + 1 = 0 If w is a root, we can see that w2 is another root. Since, w2 + w + 1 = 0, we can say that sum of cube roots of unity is zero. read less
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