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Question Number 202325 by Ikbal last updated on 24/Dec/23
Answered by deleteduser1 last updated on 24/Dec/23
Letx=reiθ=r(cosθ+isin(θ));1x=e−iθr=1r(cos(θ)−isin(θ))⇒xn=rn[cos(nθ)+isin(nθ)]1xn=1rn[cos(nθ)−isin(nθ)]x+1x=2cos(θ)⇒(r+1r)cos(θ)+(r−1r)isin(θ)⇒r+1r=2∧(r−1r)sin(θ)=0⇒r2−2r+1=0⇒r=1xn+1xn=(rn+1rn)cos(nθ)+(rn−1rn)isin(nθ)=2cos(nθ)
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