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Complex Roots

  1. Jan 8, 2009 #1
    1. Find the solution of the following

    (w)^4 = 1


    w can be a complex number (in polar form)

    w^n = re^jntheta (0 <= theta < 2pie)

    1 = 1e^j(2pie*k) k = 0, 1, 2 ,3 .........


    equating the two
    ----------------------------------------
    r = 1

    theta*n = 2pie*k
    theta = 2pie*k/n


    for k = 0 theta = 0
    for k = 1 theta = 2pie/4
    for k = 2 theta = 4pie/4
    for k = 3 theta = 6pie/4

    so there are 4 roots with magnitude 1 and the angles above.



    NOW im confused on how would I apply a similar approach to a question like the following:

    (w - (1+ j2))^5 = (32/sqrt(2))(1 + j)


    Any help appreciated!
     
    Last edited: Jan 8, 2009
  2. jcsd
  3. Jan 8, 2009 #2

    tiny-tim

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    Hi salman213! :smile:

    Hint: When the RHS had arg = 0, you wrote out the arg in all possible different ways … in that case, arg = 0, 2π, 4π, 6π … and then divided.

    Here, the RHS has arg = π/4, so all the possible ways of writing it are … ? :smile:
     
  4. Jan 8, 2009 #3
    hello tim,

    so
    (w - (1+ j2))^5 = (32/sqrt(2))(1.414)e^j45

    w = re^jtheta

    1 + j2 = (2.23)e^j63.43


    so would u say r - 2.23 = 25.82
    r= 28.05

    i think i just went off track....how to i relate the angle..:S?
     
  5. Jan 9, 2009 #4

    tiny-tim

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    Hello salman213! :smile:

    (have a square-root: √ and a pi: π :wink:)
    No, do it one step at a time …

    (and where does 2.23 come from? (32)1/5 = 2)

    take the fifth root of each side first …

    (w - (1+ j2)) = 2(ejπ/4)1/5 = one of five posiible values … :smile:
     
  6. Jan 9, 2009 #5
    but hmmm

    so one answer for w = 2(e^jπ/4)^1/5 + (1+ j2) ?

    but do i have to some how simplify this so I have w = re^jtheta ? or just leave it ?


    and for the other answers i think i would just have to add 2pie ? to the angles ? but which angle :( if that is correct?
     
  7. Jan 9, 2009 #6

    tiny-tim

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    Hello salman213! :smile:
    You can't just leave it …

    it's neither one thing nor the other :rolleyes:

    expand the (ejπ/4)1/5 as cos + jsin so that the whole thing becomes a + jb
    (ejπ/4)1/5 has 5 values …

    so you add multiples of 2π to … ? :smile:
     
  8. Jan 9, 2009 #7
    to π/4 ?
     
  9. Jan 9, 2009 #8

    tiny-tim

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    Yup! :biggrin:
     
  10. Jan 9, 2009 #9
    so the solution for w is

    w = 2(e^j(2π*k+π/4)^1/5 + (1+ j2) k = 0, 1, 2, 3, and 4 ?


    would this be always the case, if you have such solutions to find would u always take the the nth root of the RHS and then add 2pie to the angle?

    when would u add for example PIE :S?
     
  11. Jan 9, 2009 #10

    tiny-tim

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    It's clearer if you write it as

    w = 2(e^j(π/20 + (2πk/5)) + (1+ j2) k = 0, 1, 2, 3, and 4 …

    then you can see exactly where the 2π comes :smile:

    (and no, it wouldn't be π, it's always 2π :wink:)
     
  12. Jan 9, 2009 #11
    also actually one of the preliminary steps what happened to the √2

    in the original question (w - (1+ j2))^5 = (32/sqrt(2))(1 + j)

    32/√2 disappeared :(
     
  13. Jan 9, 2009 #12

    tiny-tim

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    Hint: what's ejπ/4 (in the form a + jb)? :wink:
     
  14. Jan 9, 2009 #13
    ohhkk got ya... it cancels out with the magnitude of that..

    so my final answer would be w = 2(e^j(π/20 + (2πk/5)) + (1+ j2)

    w = [2cos(π/20 + (2πk/5) + 1] + j[2sin(π/20 + (2πk/5)) + 2] k = 0, 1, 2, 3, and 4 …

    so in case i had other powers example in general

    (w - complex)^p = complex


    the first step is to

    w = complex^1/p + complex

    and the complex^1/p has an angle which u add 2pie to p times


    would you say that is correct ?
     
  15. Jan 9, 2009 #14

    tiny-tim

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    Yes! :smile:
     
  16. Jan 9, 2009 #15
    thank you tim!!!!
     
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