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Math Help - Complex numbers in polar

  1. #1
    Member classicstrings's Avatar
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    Complex numbers in polar

    Hi all! Once again stuck on complex numbers...

    (a) Expand (c+is)^4, where i^2 = -1. Hence by using De Moivre's theorem, show that

    sin4@/cos4@ = (4c^3s - 4cs^3)/ (c^4 - 6c^2s^2 + s^4), where c = cos@, and s = sin @

    I expanded to get c^4 + 4c^3si - 6c^2s^2 - 6cs^3i + s^4, and I see that sin4@/cos4@ is just s^4/c^4. But not sure how to show the solution.

    (b) By substituting t = s/c = tan@, deduce an expression for tan4@ in terms of t only. Hence show that tan4@ = 0, when 4t - 4t^3 = 0

    (c) Solve the equation obtained in part b for t. Hence find the values of @ in the range -pie/2 less than @ less than pie/2, for which tan4@ = 0.

    Thanks once again for the help!
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  2. #2
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    Hello, classicstrings!

    You're that close in part (a) . . .


    (a) Expand (cosθ + isinθ)^4, where i = -1.
    Hence by using De Moivre's theorem, show that:

    . . sin4θ/cos4θ .= .(4cs - 4cs) / (c^4 - 6cs + s^4), where c = cosθ, and s = sinθ


    I expanded to get: c^4 + 4csi - 6cs - 6csi + s^4,

    and I see that: sin4θ/cos4θ is just s^4/c^4.
    But not sure how to show the solution.

    DeMoivre's Theorem says: .(cosθ + isinθ)^4 .= .cos4θ + isin4θ

    Your expansion is: .cos4θ + isin4θ .= .(c^4 - 6cs + s^4) + i(4cs - 4cs)


    Equate real and imaginary components:

    . . cos4θ .= .c^4 - 6cs + s^4
    . . sin4θ .= .4cs - 4cs

    . . . . . . . . . sin4θ . . . . . . . 4cs - 4cs
    Therefore: . -------- . = . ----------------------
    . . . . . . . . . cos4θ . . . . c^4 - 6cs + s^4

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  3. #3
    Member classicstrings's Avatar
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    Thanks for the help on the first part.

    Still having problems with the next two parts. How do you simplify the expression, tanθ =

    4cs - 4cs
    ----------------------
    c^4 - 6cs + s^4

    After substituting c = cosθ, s = sinθ

    4cs - 4cs = 4c(s-1)
    c^4 - 6cs + s^4 = (c-3cs) - 9cs + s
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  4. #4
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    Hello again, classicstrings!

    How do you simplify the expression:

    . . . . . . . . . . . . . 4cs - 4cs
    . . tan4θ . = . ----------------------
    . . . . . . . . . .c^4 - 6cs + s^4

    Divide top and bottom by c^4:

    . . . . . . . . . . . . .4cs . . 4cs
    . . . . . . . . . . . . ------- - ------
    . . . . . . . . . . . . .c^4 . . .c^4 . . . . . . . . . 4(s/c) - 4(s/c)
    . . tan4θ . = . ------------------------ . = . -------------------------
    . . . . . . . . . . .c^4 . 6cs . . s^4 . . . . 1 - 6(s/c) + (s/c)^4
    . . . . . . . . . . .---- - ------- + -----
    . . . . . . . . . . .c^4 . .c^4 . . .c^4


    . . . . . . . . . . . . . . . . . . 4tanθ - 4tanθ
    Therefore: . tan4θ . = . -------------------------
    . . . . . . . . . . . . . . . . .1 - 6tanθ + tan^4θ

    Last edited by Soroban; March 9th 2007 at 03:49 AM.
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  5. #5
    Member classicstrings's Avatar
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    Thanks Soroban!

    So the next part, of course when 4t - 4t^3 = 0, then the top is 0, and tan4θ = 0.

    Now part c)

    4t - 4t^3 = 4t(1-6t^2+t^4)
    .
    .
    .
    4t^3(t-root5)(t+root5)

    Sub in t = tanθ,

    So, θ = 0, t+-root5 = 0, θ = 65.9 degrees, - 65.9 degrees.

    Would you please tell me if this is the right way Soroban? Thanks so much!
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  6. #6
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    Hello again, classicstrings!

    Solve the equation in part b for t.
    Hence find the value of θ in the range -π < θ < π for which tan4θ = 0.

    If tan4θ = 0, then: .4t - 4t .= .0

    We have: .4tanθ - 4tanθ .= .0

    Factor: .4tanθ (1 - tanθ) .= .0


    Solve the two equatons:

    . . 4tanθ = 0 . . tanθ = 0 . . θ = 0

    . . 1 - tanθ .= .0 . . tanθ = 1 . . tanθ = 1 . . θ = π

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