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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- May 12th 2007, 03:51 AMahoy hoycomplex numbers
ok..

w=cos(theta) +*i*sin(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. - May 12th 2007, 04:36 AMSoroban
Hello, ahoy hoy!

Quote:

w .= .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 - 2*i*·sinθ·cosθ - sin²θ - 5·cosθ + 5*i*·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

- May 12th 2007, 04:47 AMahoy hoyforever grateful
u legend..

u seem very huggable right now :D

one teency thing tho- why does

5

Note that: .---------------- . = . 5(cosθ -*i*·sinθ)

. . . . . . . . .cosθ +*i*·sinθ

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

either way, thanks heaps. - May 12th 2007, 04:52 AMahoy hoyfalse 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.