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Using z=cos(θ)+isin(θ), find expressions for z^n-1/z^n and z^n+1/z^n

We make use of De Moivre's Theorem which states that (cos(θ)+isin(θ))^n=cos(nθ)+isin(nθ).z^n-1/z^n=cos(nθ)+ isin(nθ)-cos(-nθ)- isin(-nθ)=cos(nθ)+ isin(nθ)-cos(nθ)+ isin(nθ) (by trig relationships)=2isin(nθ)Similarly z^n+1/z^n=cos(nθ)+ isin(nθ)+cos(-nθ)+isin(-nθ)=cos(nθ)+ isin(nθ)+cos(nθ)- isin(nθ) (by trig relationships)=2cos(nθ)

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Using z=cos(θ)+isin(θ), find expressions for z^n-1/z^n and z^n+1/z^n

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Using z=cos(θ)+isin(θ), find expressions for z^n-1/z^n and z^n+1/z^n

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