(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

If z = cosθ + isinθ, show that z^{n}+ z^{-n}= 2cosnθ. Hence solve the equation 3z^{4}- z^{3}+ 4z^{2}- z + 3 = 0 for complex z.

(Hint: Reduce the given equation to an equation with the unknown cosθ)

2. Relevant equations

z = cosθ + isinθ

z^{n}+ z^{-n}= 2cos(nθ)

3z^{4}- z^{3}+ 4z^{2}- z + 3 = 0

cos2θ = 2cos^{2}θ - 1

3. The attempt at a solution

From de Moivre's Theorem

z^{n}= cos(nθ) + isin(nθ)ANDz^{-n}= cos(-nθ) + isin(-nθ) = cos(nθ) - isin(nθ)

Hence z^{n}+ z^{-n}

= cos(nθ) + isin(nθ) + cos(nθ) - isin(nθ)

= 2cos(nθ)

3z^{4}- z^{3}+ 4z^{2}- z + 3 = 0

Divide both sides by z^{2}

3z^{2}- z + 4 - z^{-1}+ 3z^{-2}= 0

Rearranging to

3z^{2}+ 3z^{-2}- z - z^{-1}+ 4 = 0

3(z^{2}+ z^{-2}) - (z + z^{-1}) + 4 = 0

Using z^{n}+ z^{-n}= 2cos(nθ)

3(2cos2θ) - (2cosθ) + 4 = 0

Substituting cos2θ = 2cos^{2}θ - 1

3(2[2cos^{2}θ - 1]) - 2cosθ + 4 = 0

Simplifies down to

6cos^{2}θ - cosθ - 1 = 0

Solving for cosθ

(2cosθ - 1)(3cosθ + 1) = 0

Therefore cosθ = 1/2cosθ = -1/3OR

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)

z= 1/2 + sqrt(3)/2 i_{1}

z= 1/2 - sqrt(3)/2 i_{2}

z= -1/3 + sin[cos_{3}^{-1}(-1/3)] i

z= -1/3 - sin[cos_{4}^{-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 z_{3}and z_{4}. So my question is have I the whole question correctly? If not can you show me how it should be done. Thanks in advance!

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# Homework Help: Solving a polynomial with complex properties!

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