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. . . . . . . . . . . . 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