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Math Help - Complex Numbers

  1. #1
    Member classicstrings's Avatar
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    Complex Numbers

    Note: z = cis(@) here.

    I worked out that

    z + z^-1 = 2cos(@) ---1

    (z + z^-1)^3 = z^3 + z^-3 + 3(z+z^-1) --- 2

    Now I need to prove cos(3@) = 4cos^3(@) - 3cos(@)

    I try to sub in 2cos(@) into 2 for (z + z^-1), and fiddle around, but am unable to prove the above.

    Any suggestions? Thanks.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by classicstrings View Post
    Note: z = cis(@) here.

    I worked out that

    z + z^-1 = 2cos(@) ---1

    (z + z^-1)^3 = z^3 + z^-3 + 3(z+z^-1) --- 2

    Now I need to prove cos(3@) = 4cos^3(@) - 3cos(@)
    (z + z^-1)^3 = 8 cos^3(@) by 1.

    z^3 + z^-3 = cis(3@) + cis(-3@) = 2 cos(3@)

    3(z+z^-1) = 6 cos(@)

    So 2 becomes:

    8cos^3(@) = 2cos(3@) + 6cos(@)

    or:

    4cos^3(@) = cos(3@) + 3cos(@)

    RonL
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  3. #3
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    Hello, classicstrings!

    I could help more if I knew what the question is . . .

    What are we trying to prove?


    Note: z = cisθ here.

    I worked out that: .z + z^{-1} .= /2Ěcosθ . . . . good!
    Cube both sides: .[z + z^{-1}]│ .= .8Ěcos│θ

    And we have: .z^3 + z^{-3} + 3[z + z^{-1}] .= .8Ěcos│θ

    . . which becomes: .z^3 + z^{-3} + 3Ě2Ěcosθ .= .8Ěcos│θ

    . . and we have: .z^3 + z^{-3} .= .8Ěcos│θ - 6Ěcosθ


    Now I need to prove: .cos(3θ) .= .4Ěcos│θ - 3Ěcosθ . . why?
    This is a standard multiple-angle identity.

    To prove it, we need four more identities:
    . . cos(A + B) .= .cos(A)Ěcos(B) - sin(A)Ěsin(B)
    . . . . .sin(2A) .= .2Ěsin(A)Ěcos(A)
    - - . . cos(2A) .= .2Ěcos▓(A) - 1
    . . . . .sin▓(A) .= .1 - cos▓(A)


    We have: .cos(3θ) .= .cos(2θ + θ)
    . . . . . . . . . . . . . .= .cos(2θ)Ěcos(θ) - sin(2θ)Ěsin(θ)
    . . . . . . . . . . . . . .= .[2Ěcos▓(θ) - 1]Ěcos(θ) - [2Ěsin(θ)Ěcos(θ)]Ěsin(θ)
    . . . . . . . . . . . . . .= .2Ěcos│(θ) - cos(θ) - 2Ěcos(θ)Ěsin▓(θ)
    . . . . . . . . . . . . . .= .2Ěcos│(θ) - cos(θ) - 2Ěcos(θ)[1 - cos▓(θ)]
    . . . . . . . . . . . . . .= .2Ěcos│(θ) - cos(θ) - 2Ěcos(θ) + 2Ěcos│(θ)
    . . . . . . . . . . . . . .= .4Ěcos│(θ) - 3Ěcos(θ)

    . . . . . ta-DAA!

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  4. #4
    Member classicstrings's Avatar
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    Hello Soroban!

    The point of the question was to use the previous two results to prove the cos(3θ) .= .4Ěcos│θ - 3Ěcosθ .

    Looking over CB's solution I was able to get it.

    Now I have to consider z - z^-1, to find an expression for sin3@ in terms of sin@.

    So z - z^-1 = 2isin@

    8isin^3@ = z^3-z^-3 - 3(z - z^-1)
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  5. #5
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    Hello again, classicstrings!

    Then you were only a few steps away . . . and so was I.


    We have: .z + z^{-1} .= .2Ěcosθ

    Cube both sides: .[z + z^{-1}]│ .= .8Ěcos│θ

    And we have: .z^3 + z^{-3} + 3[z + z^{-1}] .= .8Ěcos│θ

    . . which becomes: .z^3 + z^{-3} + 3Ě2Ěcosθ .= .8Ěcos│θ

    . . and we have: .z^3 + z^{-3} .= .8Ěcos│θ - 6Ěcosθ .[1]
    Note that:
    . . z^3 . = .cos3θ + iĚsin3θ
    . . z^-3 .= .cos3θ - iĚsin3θ

    Add: . z^3 + z^{-3} .= .2Ěcos3θ .[2]


    Equate [1] and [2]: .2Ěcos3θ .= .8Ěcos│θ - 6Ěcosθ . . cos3θ .= .4Ěcos│θ - 3Ěcosθ

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