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## Homework Statement

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)

## Homework Equations

z = cosθ + isinθ

z

^{n}+ z

^{-n}= 2cos(nθ)

3z

^{4}- z

^{3}+ 4z

^{2}- z + 3 = 0

cos2θ = 2cos

^{2}θ - 1

## The Attempt at a Solution

From de Moivre's Theorem

z

^{n}= cos(nθ) + isin(nθ)

**AND**z

^{-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 = 1/2

**cosθ = -1/3**

*OR*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!