Fie z un numar complex nenul si p un numar natural nenul astfel incat \( z^p+\frac{1}{z^p} \) este un numar real din intervalul \( [-2,2] \).
Sa se arate ca \( z^n+\frac{1}{z^n} \) este de asemenea un numar real din intervalul \( [-2,2] \) pentru orice \( n \) numar natural.
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Numere complexe
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Marius Mainea
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andy crisan
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Fie \( z=r(\cos\alpha+i\sin\alpha) \) \( \Rightarrow \) \( z^{p}+\frac{1}{z^{p}}=r^{p}(\cos p\alpha+i\sin p\alpha)+\frac{1}{r^{p}}(\cos p\alpha-i\sin p\alpha)=\cos p\alpha(r^{p}+\frac{1}{r^{p}})+i\sin p\alpha(r^{p}-\frac{1}{r^{p}})\in \mathbb{R} \) \( \Rightarrow \) \( \sin p\alpha(r^{p}-\frac{1}{r^{p}})=0 \)
Cazul 1. Daca \( r^{p}-\frac{1}{r^{p}}=0\Rightarrow r=1 \Rightarrow z^{n}+\frac{1}{z^{n}}=2\cos n\alpha\in[-2,2]\forall n \in\mathbb{N} \).
Cazul 2. Daca \( \sin p\alpha=0\Rightarrow \alpha=\frac{k\pi}{p}\Rightarrow z^{p}+\frac{1}{z^{p}}= \cos k\pi(r^{p}+\frac{1}{r^{p}}) \).
Cum \( \cos k\pi\in \)\( \left{-1,1\right} \) si cum \( r^{p}+\frac{1}{r^{p}}\geq2\Rightarrow \cos k\pi(r^{p}+\frac{1}{r^{p}})\in(-\infty,-2] \) sau \( \in [2,\infty) \), dar cum \( \cos k\pi(r^{p}+\frac{1}{r^{p}})\in[-2,2] \) au loc egalitatile \( r^{p}=\frac{1}{r^{p}}\Rightarrow r=1 \). Analog ca la cazul precedent obtinem ca \( z^{n}+\frac{1}{z^{n}}\in\left[-2,2\right] \)\( \forall n \in \mathbb{N} \).
Cazul 1. Daca \( r^{p}-\frac{1}{r^{p}}=0\Rightarrow r=1 \Rightarrow z^{n}+\frac{1}{z^{n}}=2\cos n\alpha\in[-2,2]\forall n \in\mathbb{N} \).
Cazul 2. Daca \( \sin p\alpha=0\Rightarrow \alpha=\frac{k\pi}{p}\Rightarrow z^{p}+\frac{1}{z^{p}}= \cos k\pi(r^{p}+\frac{1}{r^{p}}) \).
Cum \( \cos k\pi\in \)\( \left{-1,1\right} \) si cum \( r^{p}+\frac{1}{r^{p}}\geq2\Rightarrow \cos k\pi(r^{p}+\frac{1}{r^{p}})\in(-\infty,-2] \) sau \( \in [2,\infty) \), dar cum \( \cos k\pi(r^{p}+\frac{1}{r^{p}})\in[-2,2] \) au loc egalitatile \( r^{p}=\frac{1}{r^{p}}\Rightarrow r=1 \). Analog ca la cazul precedent obtinem ca \( z^{n}+\frac{1}{z^{n}}\in\left[-2,2\right] \)\( \forall n \in \mathbb{N} \).