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The roots of the cubic equation (z + ab)...

The roots of the cubic equation `(z + ab)^(3) = a ^(3) , a ne 0` represent the vertices of a triangle of sides of length

A

`(1)/(sqrt(3)) |ab|`

B

`sqrt(3)|a|`

C

`sqrt(3)|b|`

D

`(1)/( sqrt(3))|a|`

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The correct Answer is:
To solve the given problem, we start with the cubic equation: \[ (z + ab)^3 = a^3 \] ### Step 1: Simplify the Equation We can rewrite the equation as: \[ z + ab = a \cdot \omega^k \] where \( \omega = e^{2\pi i / 3} \) is a primitive cube root of unity, and \( k = 0, 1, 2 \). This gives us three roots: \[ z_1 = a - ab \] \[ z_2 = a \cdot \omega - ab \] \[ z_3 = a \cdot \omega^2 - ab \] ### Step 2: Calculate the Length of the Sides of the Triangle We need to find the lengths of the sides of the triangle formed by these roots. The lengths of the sides can be calculated using the distance formula for complex numbers: \[ |z_i - z_j| = |(x_1 + iy_1) - (x_2 + iy_2)| = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \] #### Length of Side \( z_1 - z_2 \) Calculating \( |z_1 - z_2| \): \[ z_1 - z_2 = (a - ab) - (a \cdot \omega - ab) = a(1 - \omega) \] Now, we find the magnitude: \[ |z_1 - z_2| = |a| \cdot |1 - \omega| \] Using \( \omega = -\frac{1}{2} + i \frac{\sqrt{3}}{2} \): \[ 1 - \omega = 1 + \frac{1}{2} - i \frac{\sqrt{3}}{2} = \frac{3}{2} - i \frac{\sqrt{3}}{2} \] Calculating the magnitude: \[ |1 - \omega| = \sqrt{\left(\frac{3}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{9}{4} + \frac{3}{4}} = \sqrt{3} \] Thus, \[ |z_1 - z_2| = |a| \cdot \sqrt{3} \] #### Length of Side \( z_2 - z_3 \) Calculating \( |z_2 - z_3| \): \[ z_2 - z_3 = (a \cdot \omega - ab) - (a \cdot \omega^2 - ab) = a(\omega - \omega^2) \] Calculating the magnitude: \[ |z_2 - z_3| = |a| \cdot |\omega - \omega^2| \] Using \( \omega - \omega^2 = \omega + \frac{1}{2} - i \frac{\sqrt{3}}{2} \): \[ |\omega - \omega^2| = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \] Thus, \[ |z_2 - z_3| = |a| \cdot 1 = |a| \] #### Length of Side \( z_1 - z_3 \) Calculating \( |z_1 - z_3| \): \[ z_1 - z_3 = (a - ab) - (a \cdot \omega^2 - ab) = a(1 - \omega^2) \] Calculating the magnitude: \[ |z_1 - z_3| = |a| \cdot |1 - \omega^2| \] Using \( 1 - \omega^2 = 1 + \frac{1}{2} + i \frac{\sqrt{3}}{2} = \frac{3}{2} + i \frac{\sqrt{3}}{2} \): \[ |1 - \omega^2| = \sqrt{\left(\frac{3}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{9}{4} + \frac{3}{4}} = \sqrt{3} \] Thus, \[ |z_1 - z_3| = |a| \cdot \sqrt{3} \] ### Conclusion The lengths of the sides of the triangle are: 1. \( |z_1 - z_2| = |a| \cdot \sqrt{3} \) 2. \( |z_2 - z_3| = |a| \cdot 1 \) 3. \( |z_1 - z_3| = |a| \cdot \sqrt{3} \) Since all sides are equal, the triangle is equilateral.
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ML KHANNA-COMPLEX NUMBERS -Problem Set (2) (M.C.Q)
  1. If omega is an imaginary fifth root of unity, then find the value of l...

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  2. The product of all n^(th) root of unity is always

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  3. The roots of the cubic equation (z + ab)^(3) = a ^(3) , a ne 0 represe...

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  4. If If 1, alpha(1), alpha(2), alpha(3) , . . . alpha(n - 1) are the n...

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  5. If beta ne 1 be any nth root of unity, then 1 + 3 beta + 5 beta ^...

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  6. If 1,alpha,alpha^(2),……….,alpha^(n-1) are n^(th) root of unity, the va...

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  7. If p is a multiple of n , then the sum of pth powers of nth roots of u...

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  8. If p is not a multiple of n, then the sum of pth powers of nth root...

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  9. If alpha (1), alpha(2), . . . alpha (100) are all the 100 th roots of...

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  10. If z = ( sqrt(3) + i)/( 2) then ( z^(101) + i ^(103))^(105) equals

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  11. The square root of 3+4i is

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  12. The square root of the number 5 + 12 i is

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  13. The square root of the numbers (- 7 - 24i) is

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  14. (sqrt(5 + 12 i ) + sqrt( 5 - 12 i) ) /(sqrt(5 + 12 i ) - sqrt( 5 - 12 ...

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  15. If ( sqrt(3) + i) ^(100) = 2^(99) (a + ib) , " then " a^(2) + b^(2) i...

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  16. If ( sqrt(3) + i) ^(100) = 2 ^(99) (a + i b) then b =

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  17. If ( sqrt( 3) + i) ^(10) = a + ib then, a and b are respectively

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  18. The solution of the equation (1 + i sqrt(3))^(x) = 2^(x) are in

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  19. If (x+i y)^5=p+i q , then prove that (y+i x)^5=q+i pdot

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  20. Sum of sixth power of the roots of the equation t^(2) - 2t + 4 = 0 i...

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