0-2pi-solve-in-C-this-equation-z-2-2-1-cos-z-2-2-0-

Question Number 121684 by mathocean1 last updated on 10/Nov/20

θ \in [0; 2 π] .

s o l v e i n C t h i s e q u a t i o n :

z^{2} - (2^{θ + 1} c o s θ) z + 2^{2 θ} = 0

Commented by Dwaipayan Shikari last updated on 10/Nov/20

z = \frac{2^{θ + 1} c o s θ \pm \sqrt{2^{2 (θ + 1)} {c o s}^{2} θ - 4 . 2^{2 θ}}}{2}

z = \frac{2^{θ + 1} c o s θ \pm 2^{(θ + 1)} i s i n θ}{2}

z = 2^{θ} (c o s θ \pm i s i n θ)