Answer to Question #214419 in Complex Analysis for prince

Question

(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

Answer 1

(8.1) Let $z = \frac{z_1}{z_2}$ where $z_1 = \tan θ + i$ and $z_2 = z_1$. It follows that $z = \frac{z_1}{z_2}=\frac{\tan θ + i}{\tan θ + i}=1,$ and hence $z^n=1$ for any $n \in\N.$ .

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

$\frac{2z + i}{ −1 − iz}=\frac{2(\cos θ − i(1 +\sin θ))+i}{-1-i(\cos θ − i(1 +\sin θ))}= \frac{2\cos θ + i(-1 +2\sin θ)}{-2-\sin θ-i\cos θ}=$

$-\frac{2\cos θ + i(-1 +2\sin θ)}{2+\sin θ+i\cos θ}=-\frac{(2\cos θ + i(-1 +2\sin θ))(2+\sin θ-i\cos θ)}{(2+\sin θ+i\cos θ)(2+\sin θ-i\cos θ)}= -\frac{2\cos θ+2\cos θ\sin θ-2i\cos^2 θ+i(-2-\sin θ+4\sin θ+2\sin^2 θ)-\cos θ+2\sin θ\cos θ }{(2+\sin θ)^2+\cos^2 θ}= -\frac{\cos θ+2\sin 2θ+i(-2+3\sin θ-2\cos 2θ) }{5+4\sin θ}= -\frac{\cos θ+2\sin 2θ}{5+4\sin θ}+i\frac{2-3\sin θ+2\cos 2θ }{5+4\sin θ}.$