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Question Number 164820 by Avijit007 last updated on 22/Jan/22

Answered by Ar Brandon last updated on 22/Jan/22

$$\mathrm{Let}x\mathrm{and}y\mathrm{be}\mathrm{complex}\mathrm{numbers}\mathrm{such}\mathrm{that}x=e^{i\vartheta }\mathrm{and}y=e^{i\mathrm{\Phi }}$$$$\mathrm{Then}{x}^m+\frac{1}{x^m}=2\cos \left(m\vartheta \right)\mathrm{and}{y}^m+\frac{1}{y^m}=2\cos \left(m\mathrm{\Phi }\right)$$$$\Rightarrow (x^m+\frac{1}{x^m})(y^m+\frac{1}{y^m})=4\cos \left(m\vartheta \right)\cos \left(m\mathrm{\Phi }\right)$$$$\Rightarrow x^m{y}^m+\frac{1}{x^m{y}^m}+{\left(\frac{x}{y}\right)}^m+{\left(\frac{y}{x}\right)}^m=4\cos \left(m\vartheta \right)\cos \left(m\mathrm{\Phi }\right)$$$$\Rightarrow x^m{y}^m+\frac{1}{x^m{y}^m}=4\cos \left(m\vartheta \right)\cos \left(m\mathrm{\Phi }\right)-e^{i(\vartheta -\mathrm{\Phi })m}-\frac{1}{e^{i(\vartheta -\mathrm{\Phi })m}}$$$$\Rightarrow x^m{y}^m+\frac{1}{x^m{y}^m}=4\cos \left(m\vartheta \right)\cos \left(m\mathrm{\Phi }\right)-2\cos (m\vartheta -m\mathrm{\Phi })$$$$\Rightarrow x^m{y}^m+\frac{1}{x^m{y}^m}=2\cos \left(m\vartheta \right)\cos \left(m\mathrm{\Phi }\right)-2\sin \left(m\vartheta \right)\sin \left(m\mathrm{\Phi }\right)$$$$\Rightarrow x^m{y}^m+\frac{1}{x^m{y}^m}=2\cos (m\vartheta +m\mathrm{\Phi })$$

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