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Question Number 206007 by SANOGO last updated on 04/Apr/24

$$(E,<,>):prouve$$$$<x,y>=\frac{1}{4}\underset{k=o}{\sum ^3}{i}^k\mid \mid x+i^{k}y\mid {\mid }^2$$

Commented by Berbere last updated on 04/Apr/24

$$Wecangenerlizedit$$$$<x,y>=\frac{1}{n}\underset{k=0}{\sum ^n}{e}^{\frac{2ik\pi }{n}}\mid \mid x+e^{\frac{2ik\pi }{n}}y\mid {\mid }^2$$

Answered by Berbere last updated on 04/Apr/24

$$\mid \mid x+i^{k}y\mid {\mid }^2=<x+i^{k}y;x+i^{k}y>$$$$<x,y>=(<y\stackrel{-}{,}x>)$$$$\mathrm{\forall }(x,y)\in E^2;\mathrm{\forall }(a,b)\in \mathbb{C}$$$$<ax,by>=a\stackrel{-}{b}<x,y>;<x+x^{\prime };y>=<x,y>+<x^{\prime },y>$$$$<x,y+y^{\prime }>=<x,y>+<x,y^{\prime }>$$$$<x+i^{k}y;x+i^{k}y>$$$$=<x,x>+{(-i)}^k<x,y>+i^k<y,x>+<y,y>$$$$=\mid \mid x\mid {\mid }^2+\mid \mid y\mid {\mid }^2+{(-i)}^k<x,y>+i^k<y,x>$$$$\mathrm{\Sigma }{i}^k=0;i^{k}rootof{X}^4-1;\mathrm{\Sigma }root=0$$$$\frac{1}{4}\mathrm{\Sigma }{i}^k(\mid \mid x\mid {\mid }^2+\mid \mid y\mid {\mid }^2+{(-i)}^k<x,y>+i^k<y,x>)$$$$=\frac{1}{4}(\mid \mid x^2\mid \mid +\mid \mid y\mid {\mid }^2)\mathrm{\Sigma }{i}^k+\frac{1}{4}\underset{k=0}{\sum ^3}{i}^k{(-i)}^k<x,y>$$$$+\frac{1}{4}\underset{k=0}{\sum ^3}{\left(i\right)}^k\left(i^k\right)<y,x>$$$$=<x,y>$$$$\mathrm{\Sigma }{\left(i\right)}^k{(-i)}^k=\mathrm{\Sigma }1=4;\mathrm{\Sigma }{\left(i\right)}^k{\left(i\right)}^k=\underset{0}{\sum ^3}{(-1)}^k=0$$$$<x,y>.\frac{1}{4}\underset{k=0}{\sum ^3}{i}^k\mid \mid x+i^{k}y\mid {\mid }^2$$

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