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Math Help - complex numbers

  1. #1
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    complex numbers

    ok..
    w=cos(theta) + isin(theta) where theta is between 0 and pi.
    if complex number w^2 + (5/w) -2 is purely imaginary, show that 2(cos^2)(theta) + 5cos(theta) -3=0.
    Hence, find w.
    Any input would be appreciated. thanks.
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  2. #2
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    Hello, ahoy hoy!

    w .= .cosθ + iĚsinθ, where 0 < θ < π.

    If complex number w▓ + (5/w) - 2 is purely imaginary,
    . . show that: .2Ěcos▓θ + 5Ěcosθ - 3 .= .0.

    Hence, find w.
    . . . . . . . . . . . . 5
    Note that: .---------------- . = . 5(cosθ - iĚsinθ)
    . . . . . . . . .cosθ + iĚsinθ


    We have: .(cosθ + iĚsinθ)▓ - 5(cosθ - iĚsinθ) - 2

    . . = .cos▓θ - 2iĚsinθĚcosθ - sin▓θ - 5Ěcosθ + 5iĚsinθ - 2

    . . = .(cos▓θ - sin▓θ + 5Ěcosθ - 2) + i(2ĚsinθĚcosθ - 5Ěsinθ)

    Since this quantity is pure imaginary, its real component is zero.
    . . Hence: .cos▓θ - sin▓θ + 5Ěcosθ - 2 .= .0

    We have: .cos▓θ - (1 - cos▓θ) + 5Ěcosθ - 2 .= .0

    Therefore: .2Ěcos▓θ + 5Ěcosθ - 3 .= .0


    This equation factors: .(cosθ + 3)(2Ěcosθ - 1) .= .0

    . . Then: .cosθ + 3 .= .0 . . . . cosθ = -3 . . . . no real roots

    . . And: .2Ěcosθ - 1 .= .0 . . . . cosθ = Ż . . . . θ = π/3


    Therefore: .w .= .cos(π/3) + iĚsin(π/3) . . . . w .= .Ż + iĚŻ√3

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  3. #3
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    forever grateful

    u legend..
    u seem very huggable right now
    one teency thing tho- why does
    5
    Note that: .---------------- . = . 5(cosθ - iĚsinθ)
    . . . . . . . . .cosθ + iĚsinθ


    i know its 5(cos theta + i sin theta)^-1

    either way, thanks heaps.
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  4. #4
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    false alarm

    wait...is it cos u times by the conjugate of denominator, cos that seems to be working. ye nvm. thank u to the moon and back.
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