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

$$\theta \in [0;2\pi ].$$$$solvein\mathbb{C}thisequation:$$$$z^2-\left(2^{\theta +1}cos\theta \right)z+2^{2\theta }=0$$$$$$

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

$$\mathit{z}=\frac{2^{\mathit{\theta }+1}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\pm \sqrt{2^{2(\mathit{\theta }+1)}{\mathit{c}\mathit{o}\mathit{s}}^2\mathit{\theta }-4.2^{2\theta }}}{2}$$$$\mathit{z}=\frac{2^{\mathit{\theta }+1}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\pm 2^{(\theta +1)}\mathit{i}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }}{2}$$$$\mathit{z}=2^{\mathit{\theta }}(\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\pm \mathit{i}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta })$$

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