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jkasdhk

If α and β are complex cube roots of unity then (α^4 β^4)+1/αβ=?

  • one year ago
  • one year ago

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  1. matricked
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    is it 2

    • one year ago
  2. jkasdhk
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    How?

    • one year ago
  3. matricked
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    the product of the roots αβ=1 and (α^3 β^3)=1

    • one year ago
  4. jkasdhk
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    How is (α^3 β^3)=1?

    • one year ago
  5. jkasdhk
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    We see sum of roots and product of roots?

    • one year ago
  6. amrit110
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    \[\alpha * \beta \] isnt equal to 1....cube root remember?

    • one year ago
  7. jkasdhk
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    the answer given is 0. i dont know how?

    • one year ago
  8. mayank_mak
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    infact the answer is zero

    • one year ago
  9. jkasdhk
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    How????

    • one year ago
  10. amrit110
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    well its complex cube root, \[\alpha = \beta \] = \[\sqrt{-1}\]. so that would make cube of it equal to 1. but the other term -1. 1-1=0....

    • one year ago
  11. jkasdhk
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    complex cube roots means α=β?

    • one year ago
  12. mayank_mak
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    do u agree the roots \[\alpha =\Omega and \beta = \Omega ^{2}\]? or vice versa

    • one year ago
  13. amrit110
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    no, not necessarily all the time. for unity yes.

    • one year ago
  14. matricked
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    if α and β are the complex cube roots of 1 then α and β must satisfy the eq. x^2 +x +1=0 and x^3=1 thus α + β=-1 and αβ=1

    • one year ago
  15. eliassaab
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    \[ \alpha=e^{\frac{2 i \pi }{3}}; \quad \alpha^3 =1\\ \beta =1; \quad \beta^3 =1\\ \alpha \beta \ne 1 \]

    • one year ago
  16. jkasdhk
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    @amrit110: can u please tell me the steps to solve this ques?

    • one year ago
  17. mayank_mak
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    \[\Omega ^{3} = 1\]

    • one year ago
  18. amrit110
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    ok. see complex cube roots of unity so \[x^{3}=1\]. so when u solve for x it is obviously the root of (-1). now this means that for this case we can safely assume alpha n beta to be equal to x. although this is not true for all cubic equations, sometimes the roots can be complex conjugate too. but it doesnt matter.

    • one year ago
  19. shubhamsrg
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    note that the complex cube roots of unity are the zeroes of the eqn x^2 + x +1 so (alpha)(beta) = 1 try now..

    • one year ago
  20. eliassaab
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    There are 3 complex roots of unity: \[ \left\{1,e^{\frac{2 i \pi }{3}},e^{-\frac{2 i \pi }{3}}\right\} \]

    • one year ago
  21. eliassaab
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    A complex root of unity is a root of \[ x^3 =1 \]

    • one year ago
  22. matricked
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    @eliassaab i didn't get how come αβ not 1

    • one year ago
  23. jkasdhk
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    @amrit110: What will be the answer of (α^4 β^4) ?

    • one year ago
  24. eliassaab
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    If both \( \alpha \ne 1 \) and \( \beta \ne 1 \) then \(\alpha \beta =1\)

    • one year ago
  25. eliassaab
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    In the above case the ratio is 2.

    • one year ago
  26. amrit110
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    it should be 1 jkasdhk

    • one year ago
  27. matricked
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    yes for complex nos .. for real nos If both α≠1 and β≠1 then αβ not 1

    • one year ago
  28. jkasdhk
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    and 1/αβ would be?

    • one year ago
  29. matricked
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    1/αβ would be 1

    • one year ago
  30. jkasdhk
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    1+1 will be 2? right?

    • one year ago
  31. jkasdhk
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    but 2 is not the answer

    • one year ago
  32. matricked
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    is the question (α^4 β^4) -1/αβ or still different

    • one year ago
  33. eliassaab
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    If we consider complex roots only and \(\alpha \ne \beta\) then \[ \alpha = e^{\frac{2 i \pi }{3}}\\ \beta = e^{-\frac{2 i \pi }{3}}\\ \alpha \beta= e^0 =1 \]

    • one year ago
  34. jkasdhk
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    its (α^4 β^4) +1/αβ

    • one year ago
  35. eliassaab
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    If the question is: Let \( \alpha, \beta\) be the 2 complex roots (not real), then the ratio in question is 2

    • one year ago
  36. jkasdhk
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    @amrit110: How is (α^4 β^4)=1

    • one year ago
  37. eliassaab
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    \[ \alpha^4 \beta^4 = \alpha^3 \beta^3 \alpha \beta = (1)(1) =\alpha \beta =1 \] If \( \alpha\ne \beta \)

    • one year ago
  38. eliassaab
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    Let \( 1, \alpha, \beta \) be the three roots of unity,then \[ \frac { \alpha ^4 \beta^4 +1}{\alpha \beta }= 2 \] It is easy to see now \[ (1) \alpha \beta =1 \] Since the last product is the product of the three roots of the polynomial \[ x^3 -1=0

    • one year ago
  39. eliassaab
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    This product is \[ (-1)^3 \frac {-1}{1}=1 \] See http://en.wikipedia.org/wiki/Vieta%27s_formulas

    • one year ago
  40. eliassaab
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    @jkasdhk

    • one year ago
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