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anonymous

  • 5 years ago

a) Find all fourth roots of 1 in polar form. b) Express them in Cartesian form. c) Show how they can be expressed as powers of one fixed fourth root of 1.

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  1. amistre64
    • 5 years ago
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    4th roots in polar form divide the unit circle into 4 equal parts.... so 90 degree seperations

  2. amistre64
    • 5 years ago
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    since the any root of 1 = 1; im assuming they want (1,1) (1,-1) (-1,-1) (-1,1) as answers

  3. anonymous
    • 5 years ago
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    isnt it x^4=1?

  4. amistre64
    • 5 years ago
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    I was thinking \(\sqrt[4]{x}\) at \(x=1\)

  5. anonymous
    • 5 years ago
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    but i guess it is x=1^(1/4) this is what we call four forth root of 1

  6. anonymous
    • 5 years ago
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    x=1,-1,i,-i

  7. amistre64
    • 5 years ago
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    90s or the 45s....

  8. amistre64
    • 5 years ago
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    if we go with the i stuff; its the 90s and thats prolly the better interpretatio

  9. anonymous
    • 5 years ago
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    yup

  10. anonymous
    • 5 years ago
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    but for 1 its 0

  11. anonymous
    • 5 years ago
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    for -1, pi

  12. amistre64
    • 5 years ago
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    there is no pi in the cartesian; just your 1s and 0s for your intercepts

  13. amistre64
    • 5 years ago
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    (1,0) (0,1) (-1,0) (0,-1)

  14. amistre64
    • 5 years ago
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    but what 'c' is asking for I dunno

  15. amistre64
    • 5 years ago
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    maybe \((1-0i)^{1/4}\) ?

  16. amistre64
    • 5 years ago
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    or is it simply \(i^4\)

  17. anonymous
    • 5 years ago
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    i cant guess any idea

  18. anonymous
    • 5 years ago
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    i^4 seems better

  19. amistre64
    • 5 years ago
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    im thinking the first since that implies a complex plane and 4 roots

  20. anonymous
    • 5 years ago
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    but how (-1,0) and(0,-1)

  21. amistre64
    • 5 years ago
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    \((1+0i)^4\) maybe? if forget if its ^4 or ^(1/4) that pops out 4 times

  22. amistre64
    • 5 years ago
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    \(sqrt{-6}\) has complex roots right?

  23. anonymous
    • 5 years ago
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    look...when we talk about the cube roots of 1 , how we express it? x=1^1/3..no?

  24. amistre64
    • 5 years ago
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    i believe so

  25. anonymous
    • 5 years ago
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    n for forth root it is x=1^1/4

  26. anonymous
    • 5 years ago
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    \[x^4 = 1\]

  27. amistre64
    • 5 years ago
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    sqrt(-9) = 3i and we can find both those roots in the complex plane ... gonna have to dbl chk with the wolfram :)

  28. anonymous
    • 5 years ago
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    \[(x^2-1)(x^2+1)=0\]

  29. anonymous
    • 5 years ago
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    x=1,-1,i,-i

  30. anonymous
    • 5 years ago
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    for x=1 r=1 and theta =0

  31. anonymous
    • 5 years ago
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    x=-1 r=1, theta =pi -1=cospi

  32. anonymous
    • 5 years ago
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    4th roots of 1 are 1, -1, i, -1

  33. anonymous
    • 5 years ago
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    you know one answer is 1. divide unit circle (in complex plane) into 4 equal parts and you will see i, -1, -i

  34. anonymous
    • 5 years ago
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    did u get part c?

  35. anonymous
    • 5 years ago
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    what is part c?

  36. anonymous
    • 5 years ago
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    read the post:P

  37. anonymous
    • 5 years ago
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    oh yes they are all powers of i

  38. anonymous
    • 5 years ago
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    i, i^2, i^3, i^4 finito

  39. anonymous
    • 5 years ago
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    no question says" Show how they can be expressed as powers of one fixed fourth root of 1."

  40. anonymous
    • 5 years ago
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    that fixed root is i.

  41. anonymous
    • 5 years ago
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    you cannot express i as a power of 1. i assume they mean integral powers

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