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anonymous

  • 2 years ago

if z is the non real fifth root of unity then the value of ^(l 1+z+z^2+z^-2+z^-1l)=?

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  1. anonymous
    • 2 years ago
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    its find the value of 2^(l1+z+z^2+z^-2-z^-1l) sry i wrote it wrong

  2. goformit100
    • 2 years ago
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    is it IIT question ?

  3. anonymous
    • 2 years ago
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    yup

  4. anonymous
    • 2 years ago
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    you can use z+z^2+z^3+z^4=0 for z=fifth root of unity

  5. goformit100
    • 2 years ago
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    Show your working.

  6. hartnn
    • 2 years ago
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    so you know how to solve it already

  7. anonymous
    • 2 years ago
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    |dw:1389460828114:dw| what next

  8. anonymous
    • 2 years ago
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    the options are 4,2,1,none

  9. hartnn
    • 2 years ago
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    simplify this 1+z+z^2+z^-2+z^-1

  10. hartnn
    • 2 years ago
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    (????)/z^2

  11. anonymous
    • 2 years ago
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    i did it by taking the denominator |dw:1389461042399:dw|

  12. anonymous
    • 2 years ago
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    |dw:1389461100403:dw|

  13. hartnn
    • 2 years ago
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    z+z^2+z^3+z^4=0 is incorrect

  14. hartnn
    • 2 years ago
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    that gives one root as z=0

  15. hartnn
    • 2 years ago
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    z^5 =1 z^5 -1^5 = 0

  16. hartnn
    • 2 years ago
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    (z-1)(1+z+z^2+z^3+z^4) = 0 as z not = 1 (1+z+z^2+z^3+z^4) = 0

  17. hartnn
    • 2 years ago
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    so your entire numerator of the exponent goes to 0

  18. hartnn
    • 2 years ago
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    did u follow ? if yes, i think, you'll get the answer now :)

  19. anonymous
    • 2 years ago
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    no u didnt saw that i corrected the question its 1-z+z^2+z^3+z^4 and not 1+z+z^2+z^3+z^4

  20. hartnn
    • 2 years ago
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    oops

  21. anonymous
    • 2 years ago
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    |dw:1389461448241:dw|

  22. hartnn
    • 2 years ago
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    then i should say, there's typing error in the question OR the answer is NONE

  23. anonymous
    • 2 years ago
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    yup i also thought so thanks for your time by the way

  24. hartnn
    • 2 years ago
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    because z is complex, so will be 2^(-2/z) welcome ^_^

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