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anonymous

  • 5 years ago

If z being a complex number lies on the unit circle and the sum of z with its reciprocal is 1, then find z^235

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  1. anonymous
    • 5 years ago
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    wt bt this??

  2. amistre64
    • 5 years ago
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    z^235? thats alot

  3. anonymous
    • 5 years ago
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    theres no answer given bt frm what i know it comes out to be a simple answer

  4. amistre64
    • 5 years ago
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    z + 1/z = 1?

  5. anonymous
    • 5 years ago
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    yeah

  6. amistre64
    • 5 years ago
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    z^2 +1 ------- = 1; z^2 +1 = z; z^2 -2 +1 = 0 Z

  7. amistre64
    • 5 years ago
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    (z-1)^2; z = 1

  8. amistre64
    • 5 years ago
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    z^235 = 1 then ?

  9. anonymous
    • 5 years ago
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    but z isnt 1, its a complex number...

  10. amistre64
    • 5 years ago
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    typoed it again

  11. anonymous
    • 5 years ago
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    z is the 3rd root of -1. because z^2-Z+1=0; z^3=1 z^235=z=-1+isqrt(3) /2

  12. amistre64
    • 5 years ago
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    z^2+1 = 1 z^2 = 0 z=0 ?

  13. anonymous
    • 5 years ago
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    Z^3=-1 (z+1)(z^2-z+1)=0 Z^2-z+1=0;

  14. amistre64
    • 5 years ago
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    so that means the the complex part is either up or down

  15. anonymous
    • 5 years ago
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    z+1/z=1 z^2-z+1=0.

  16. anonymous
    • 5 years ago
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    i think dipankars right, did u apply the demoivres theorem??

  17. anonymous
    • 5 years ago
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    no need to apply it....

  18. amistre64
    • 5 years ago
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    dip is usually right ;)

  19. anonymous
    • 5 years ago
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    i have given the steps....

  20. anonymous
    • 5 years ago
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    z is the complex roots of -1.

  21. anonymous
    • 5 years ago
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    so but then howdu get z^235??

  22. anonymous
    • 5 years ago
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    u must realize amistre m doing wht u often do;)

  23. amistre64
    • 5 years ago
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    guess wildly lol

  24. anonymous
    • 5 years ago
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    z^235=z^234*z=(z^3*78)^z=1*z=z

  25. anonymous
    • 5 years ago
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    no not that...like wht u were doin with the tangent plane question...

  26. anonymous
    • 5 years ago
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    ok dipankar....

  27. amistre64
    • 5 years ago
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    ahhh... :)

  28. anonymous
    • 5 years ago
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    with all the three ;)

  29. amistre64
    • 5 years ago
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    its good to be smart :)

  30. anonymous
    • 5 years ago
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    him i think u have understood the process...

  31. anonymous
    • 5 years ago
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    yeah right...

  32. amistre64
    • 5 years ago
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    so, when I find z=0, the complex parts have to be pure imaginaries then right? like "i"

  33. anonymous
    • 5 years ago
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    ill tell u what

  34. anonymous
    • 5 years ago
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    z + 1/z =1 let z=cosθ + isinθ 1/z = cosθ - isinθ z + 1/z = 2cosθ = 1 cosθ = 1/2 hence z = 1/2 + √3/2

  35. amistre64
    • 5 years ago
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    z^235 = i^235...235/4 = 58 and 3 remaining i, -1, -1 z^235 = -i?

  36. anonymous
    • 5 years ago
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    √3/2 x i sorry

  37. anonymous
    • 5 years ago
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    no man look above z = 1/2 + i√3/2

  38. amistre64
    • 5 years ago
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    lol...stab in the dark then :) i see what you did

  39. anonymous
    • 5 years ago
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    now du wanna know the other one as well?

  40. amistre64
    • 5 years ago
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    i do, but at this late hour it aint gonna stick to well ;)

  41. anonymous
    • 5 years ago
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    hahaha...wts d time there??

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