Answer to Question #214419 in Complex Analysis for prince

Question #214419

(8.1) Let z = z1/z2 where z1 = tan θ + i and z2 = z1. Find an expression for z n with n ∈ N.

(8.2) Let z = cos θ − i(1 + sin θ). Determine 2z + i / −1 − iz


1
Expert's answer
2021-07-12T10:00:15-0400

(8.1) Let z=z1z2z = \frac{z_1}{z_2} where z1=tanθ+iz_1 = \tan θ + i and z2=z1z_2 = z_1. It follows that z=z1z2=tanθ+itanθ+i=1,z = \frac{z_1}{z_2}=\frac{\tan θ + i}{\tan θ + i}=1, and hence zn=1z^n=1 for any nN.n \in\N. .


(8.2) Let z=cosθi(1+sinθ)z = \cos θ − i(1 +\sin θ). Then

2z+i1iz=2(cosθi(1+sinθ))+i1i(cosθi(1+sinθ))=2cosθ+i(1+2sinθ)2sinθicosθ=\frac{2z + i}{ −1 − iz}=\frac{2(\cos θ − i(1 +\sin θ))+i}{-1-i(\cos θ − i(1 +\sin θ))}= \frac{2\cos θ + i(-1 +2\sin θ)}{-2-\sin θ-i\cos θ}=

2cosθ+i(1+2sinθ)2+sinθ+icosθ=(2cosθ+i(1+2sinθ))(2+sinθicosθ)(2+sinθ+icosθ)(2+sinθicosθ)=2cosθ+2cosθsinθ2icos2θ+i(2sinθ+4sinθ+2sin2θ)cosθ+2sinθcosθ(2+sinθ)2+cos2θ=cosθ+2sin2θ+i(2+3sinθ2cos2θ)5+4sinθ=cosθ+2sin2θ5+4sinθ+i23sinθ+2cos2θ5+4sinθ.-\frac{2\cos θ + i(-1 +2\sin θ)}{2+\sin θ+i\cos θ}=-\frac{(2\cos θ + i(-1 +2\sin θ))(2+\sin θ-i\cos θ)}{(2+\sin θ+i\cos θ)(2+\sin θ-i\cos θ)}= -\frac{2\cos θ+2\cos θ\sin θ-2i\cos^2 θ+i(-2-\sin θ+4\sin θ+2\sin^2 θ)-\cos θ+2\sin θ\cos θ }{(2+\sin θ)^2+\cos^2 θ}= -\frac{\cos θ+2\sin 2θ+i(-2+3\sin θ-2\cos 2θ) }{5+4\sin θ}= -\frac{\cos θ+2\sin 2θ}{5+4\sin θ}+i\frac{2-3\sin θ+2\cos 2θ }{5+4\sin θ}.



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