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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.
Hello, ahoy hoy!
. . . . . . . . . . . . 5w .= .cosθ + i·sinθ, where 0 < θ < π.
If complex number w² + (5/w) - 2 is purely imaginary,
. . show that: .2·cos²θ + 5·cosθ - 3 .= .0.
Hence, find w.
Note that: .---------------- . = . 5(cosθ - i·sinθ)
. . . . . . . . .cosθ + i·sinθ
We have: .(cosθ + i·sinθ)² - 5(cosθ - i·sinθ) - 2
. . = .cos²θ - 2i·sinθ·cosθ - sin²θ - 5·cosθ + 5i·sinθ - 2
. . = .(cos²θ - sin²θ + 5·cosθ - 2) + i(2·sinθ·cosθ - 5·sinθ)
Since this quantity is pure imaginary, its real component is zero.
. . Hence: .cos²θ - sin²θ + 5·cosθ - 2 .= .0
We have: .cos²θ - (1 - cos²θ) + 5·cosθ - 2 .= .0
Therefore: .2·cos²θ + 5·cosθ - 3 .= .0
This equation factors: .(cosθ + 3)(2·cosθ - 1) .= .0
. . Then: .cosθ + 3 .= .0 . . → . . cosθ = -3 . . . . no real roots
. . And: .2·cosθ - 1 .= .0 . . → . . cosθ = ½ . . → . . θ = π/3
Therefore: .w .= .cos(π/3) + i·sin(π/3) . . → . . w .= .½ + i·½√3