Show that Im(eiθ1−xeiθ)=sin⁡(θ)x2−2xcos⁡(θ)+1

zakinutuzi

zakinutuzi

Answered

2022-01-17

Show that Im(eiθ1xeiθ)=sin(θ)x22xcos(θ)+1

Answer & Explanation

Anzante2m

Anzante2m

Expert

2022-01-18Added 34 answers

Note that
eiθ1xeiθ=cos(θ)+isin(θ)1xcos(θ)ξsin(θ)
=(cos(θ)+isin(θ))(1xcos(θ)+ξsin(θ))(1xcos(θ))2+x2sin2(θ)
=cos(θ)x+isin(θ)x22xcos(θ)+1

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