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Homework Help: Express Imaginary number in standard form.

  1. Apr 23, 2007 #1
    1. The problem statement, all variables and given/known data

    Need to see if what im doing is right, here is the problem:

    z= 4i/(-1+i), express z^20 in standard form.

    so,

    z= 4i/(-1+i) = 2-2i

    z^20 = 2*sqrt(2) x cis[20(-pi/4)] = 2*sqrt(2) x cis(-5pi)
    = 2*sqrt(2) x [cos(-5pi) + i sin(-5pi)]
    = 2*sqrt(2) x [-1 + 0i]
    = 2*sqrt(2) x [-1] = -2048

    so the final answer is only a real number of magnitude -2048+0i. is this correct? thanks.
     
    Last edited: Apr 24, 2007
  2. jcsd
  3. Apr 23, 2007 #2

    Dick

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    Nope. Try again. cos(-5pi)=-1 and the magnitude of z is 2*sqrt(2). You can do better.
     
  4. Apr 24, 2007 #3
    how about now?
     
  5. Apr 24, 2007 #4

    mezarashi

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    Okay, how exactly it it get from

    2*sqrt(2) x [-1] to this number: -2048
     
  6. Apr 24, 2007 #5
    it shouldn't, thats a mistake, I was following an example with a negative exponent thus explaining why i thought it should be raised to the positive 20. but, z^20 does have to be change to z^1, so the equation will then go from this (i think):

    z^20 = [2*sqrt(2)]^(20) x cis[(20)(-pi/4)]

    to this:
    z = [2*sqrt(2)]^(1/20) x cis[(1/20)(-pi/4)]

    is that right?
     
    Last edited: Apr 24, 2007
  7. Apr 24, 2007 #6
    wait, looking at my above post, i see where i have gone wrong, I'll try to correct it...
     
  8. Apr 24, 2007 #7
    z = [2*sqrt(2)] x cis[(-pi/4)]
    = [2*sqrt(2)] * [cos(-pi/4) + i sin(-pi/4)]
    = [2*sqrt(2)] * [sqrt(2)/2 + i -sqrt(2)/2]
    = 2 - 2i

    so the answer is z^20 = 2 - 2i in standard form. is that right? (doesnt seem right, thats what i started with...)
     
    Last edited: Apr 24, 2007
  9. Apr 24, 2007 #8

    mezarashi

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    I think you're confusing yourself and confusing me as well. Are you trying to find z^(20) or z^(-20)?

    I was previously only highlighting that [tex](2 \sqrt{2})^{20}[/tex] is NOT equal to 2048.

    Here's a good old wiki page to refresh your memory:
    http://en.wikipedia.org/wiki/De_Moivre's_formula#Applications
     
    Last edited: Apr 24, 2007
  10. Apr 24, 2007 #9
    Yes, now i am confused, but the question is right:

    z= 4i/(-1+i), express z^20 in standard form.

    lets try again,

    we are raising everything in z to the power of 20,

    z^20 = [2*sqrt(2)]^20 x cis[20(-pi/4)] = [2*sqrt(2)]^20 x cis(-5pi)

    (note how large the number [2*sqrt(2)]^20, there is no reasonable way to write it other than how i have just now, this seem odd...)

    now,

    [2*sqrt(2)]^20 x cis(-5pi) = [2*sqrt(2)]^20 x [cos(-5pi)+i sin(-5pi)]

    is this correct?...

    next,

    [2*sqrt(2)]^20 x [cos(-5pi)+i sin(-5pi)] = [2*sqrt(2)]^20 x [1 + 0i]

    is this correct?...
     
    Last edited: Apr 24, 2007
  11. Apr 24, 2007 #10

    mezarashi

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    I agree with you up to:

    z^20 = [2*sqrt(2)]^20 x [cos(-5pi)+i sin(-5pi)]

    From here, it should be easy to see that cos(-5pi) = -1 and sin(-5pi) = 0. Problem solved. Note that [2*sqrt(2)]^20 = 2^30. Remember that one integer in our computer uses 4 bytes, that's 2^32.
     
  12. Apr 24, 2007 #11
    :cool: thanks.
     
  13. Apr 24, 2007 #12

    HallsofIvy

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    No, this should be (2*sqrt(2))^20 cis(-5pi)
    [/quote] = 2*sqrt(2) x [cos(-5pi) + i sin(-5pi)]
    = 2*sqrt(2) x [-1 + 0i]
    = 2*sqrt(2) x [-1] = -2048
     
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