(8.1) Let z=z2z1 where z1=tanθ+i and z2=z1. It follows that z=z2z1=tanθ+itanθ+i=1, and hence zn=1 for any n∈N. .
(8.2) Let z=cosθ−i(1+sinθ). Then
−1−iz2z+i=−1−i(cosθ−i(1+sinθ))2(cosθ−i(1+sinθ))+i=−2−sinθ−icosθ2cosθ+i(−1+2sinθ)=
−2+sinθ+icosθ2cosθ+i(−1+2sinθ)=−(2+sinθ+icosθ)(2+sinθ−icosθ)(2cosθ+i(−1+2sinθ))(2+sinθ−icosθ)=−(2+sinθ)2+cos2θ2cosθ+2cosθsinθ−2icos2θ+i(−2−sinθ+4sinθ+2sin2θ)−cosθ+2sinθcosθ=−5+4sinθcosθ+2sin2θ+i(−2+3sinθ−2cos2θ)=−5+4sinθcosθ+2sin2θ+i5+4sinθ2−3sinθ+2cos2θ.
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