Solving a polynomial with complex properties

[moocav]

Homework Statement

If z = cosθ + isinθ, show that zⁿ + z^-n = 2cosnθ. Hence solve the equation 3z⁴ - z³ + 4z² - z + 3 = 0 for complex z.
(Hint: Reduce the given equation to an equation with the unknown cosθ)

Homework Equations

z = cosθ + isinθ
zⁿ + z^-n = 2cos(nθ)
3z⁴ - z³ + 4z² - z + 3 = 0
cos2θ = 2cos²θ - 1

The Attempt at a Solution

From de Moivre's Theorem
zⁿ = cos(nθ) + isin(nθ) AND z^-n = cos(-nθ) + isin(-nθ) = cos(nθ) - isin(nθ)
Hence zⁿ + z^-n
= cos(nθ) + isin(nθ) + cos(nθ) - isin(nθ)
= 2cos(nθ)

3z⁴ - z³ + 4z² - z + 3 = 0
Divide both sides by z²
3z² - z + 4 - z^-1 + 3z^-2 = 0

Rearranging to
3z² + 3z^-2 - z - z^-1 + 4 = 0
3(z² + z^-2) - (z + z^-1) + 4 = 0

Using zⁿ + z^-n = 2cos(nθ)
3(2cos2θ) - (2cosθ) + 4 = 0

Substituting cos2θ = 2cos²θ - 1
3(2[2cos²θ - 1]) - 2cosθ + 4 = 0

Simplifies down to
6cos²θ - cosθ - 1 = 0

Solving for cosθ
(2cosθ - 1)(3cosθ + 1) = 0
Therefore cosθ = 1/2 OR cosθ = -1/3
cosθ = 1/2, θ = π/3 or 5π/3
cosθ = -1/3, θ = cos^-1(-1/3)
(Note: π = funny looking pi)

Hence (using complex conjugate is also a root of polynomial)
z₁ = 1/2 + sqrt(3)/2 i
z₂ = 1/2 - sqrt(3)/2 i
z₃ = -1/3 + sin[cos^-1(-1/3)] i
z₄ = -1/3 - sin[cos^-1(-1/3)] i

I'm pretty confident that I've done it right until the cosθ = -1/3 and getting sin[cos^-1(-1/3)] in the roots z₃ and z₄. So my question is have I the whole question correctly? If not can you show me how it should be done. Thanks in advance!

 

[vela]

You did it correctly, though you should simplify sin[arccos(-1/3)]. You can always check your answers by plugging them back into the original equation.

 

[moocav]

Ok thankyou (:

umm another question unrelated
but now I have another concern

for a function to be STRICTLY increasing, is the point 0 actually included? Because me and my friends all have different opinions. So the gradient at x = 0 is 0 so would the function still be considered as strictly increasing if it included x=0?

 

[vela]

Strictly increasing means f'(x)>0 for all x. If you have f'(x)≥0, the function is monotonically increasing.

 

[rock.freak667]

Just to add one point to your original post, when I did questions like these in Further Math, it was always best to show that z=0 was not a root before dividing by any power of z.