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Question Number 67234 by prof Abdo imad last updated on 24/Aug/19

$$factorisep\left(x\right)=1+x+x^2+x^3+x^5$$$$insideC\left[x\right]andR\left[x\right]$$$$calculatep\left(e^{i\frac{\pi }{5}}\right)andp\left(cos\left(\frac{\pi }{5}\right)\right)$$

Commented by mathmax by abdo last updated on 24/Aug/19

$$sorryp\left(x\right)=1+x+x^2+x^3+x^4$$

Commented by mathmax by abdo last updated on 25/Aug/19

$$1)p\left(x\right)=0\iff 1+x+x^2+x^3+x^4=0\iff 1-x^5=0andx\ne 1$$$$x^5=1letx={re}^{i\theta }\left(e\right)\Rightarrow r^5{e}^{i5\theta }=e^{i\left(2k\pi \right)}\Rightarrow r=1and5\theta =2k\pi \Rightarrow$$$${\theta }_k=\frac{2k\pi }{5}sotherootsare{z}_k=e^{i\left(\frac{2k\pi }{5}\right)}withk\in \left[[1,4]\right]$$$$p\left(x\right)=\prod _{k=1}^4(x-z_k)=\prod _{k=1}^4(x-e^{i\frac{2k\pi }{5}})=(x-e^{i\frac{2\pi }{5}})(x-e^{i\frac{4\pi }{5}})(x-e^{i\frac{6\pi }{5}})(x-e^{i\frac{8\pi }{5}})$$$$but{e}^{i\frac{8\pi }{5}}=e^{i(2\pi -\frac{2\pi }{5})}=e^{-\frac{i2\pi }{5}}and{e}^{i\frac{6\pi }{5}}=e^{-i\frac{4\pi }{5}}\Rightarrow$$$$p\left(x\right)=(x-e^{i\frac{2\pi }{5}})(x-e^{-i\frac{2\pi }{5}})(x-e^{i\frac{4\pi }{5}})(x-e^{-i\frac{4\pi }{5}})$$$$=(x^2-2Re\left(e^{i\frac{2\pi }{5}}\right)x+1)(x^2-2Re\left(e^{i\frac{4\pi }{5}}\right)x+1)isthedecomposition$$$$insideC\left[x\right]$$$$wehavecos\left(\frac{\pi }{5}\right)=\frac{1+\sqrt{5}}{4}\Rightarrow cos\left(\frac{2\pi }{5}\right)=2{cos}^2\left(\frac{\pi }{5}\right)-1$$$$=2{\left(\frac{1+\sqrt{5}}{4}\right)}^2-1=\frac{1}{8}(6+2\sqrt{5})-1=\frac{6+2\sqrt{5}-8}{8}=\frac{-2+2\sqrt{5}}{8}=\frac{\sqrt{5}-1}{4}$$$$cos\left(\frac{4\pi }{5}\right)=cos(\pi -\frac{\pi }{5})=-cos\left(\frac{\pi }{5}\right)\Rightarrow$$$$p\left(x\right)=(x^2-\frac{\sqrt{5}-1}{2}x+1)(x^2+\frac{\sqrt{5}-1}{2}x+1)$$

Commented by mathmax by abdo last updated on 25/Aug/19

$$wehaveforx\ne 1p\left(x\right)=\frac{1-x^5}{1-x}\Rightarrow p\left(e^{i\frac{\pi }{5}}\right)=\frac{1-{\left(e^{\frac{i\pi }{5}}\right)}^5}{1-e^{i\frac{\pi }{5}}}$$$$=\frac{2}{1-e^{\frac{i\pi }{5}}}alsowehavep\left(cos\left(\frac{\pi }{5}\right)\right)=\frac{1-{cos}^5\left(\frac{\pi }{5}\right)}{1-cos\left(\frac{\pi }{5}\right)}$$$$=\frac{1-{\left(\frac{1+\sqrt{5}}{4}\right)}^5}{1-\frac{1+\sqrt{5}}{4}}$$

Answered by mind is power last updated on 24/Aug/19

$$p\left(x\right)=1+x+x^2+x^3+x^4?$$$$$$$$$$

Answered by MJS last updated on 25/Aug/19

$$(x^4+x^3+x^2+x+1)(x-1)=x^5-1$$$$x^5=1$$$$x_0=1$$$$x_1=-\frac{1-\sqrt{5}}{4}+\frac{\sqrt{10+2\sqrt{5}}}{4}\mathrm{i}$$$$x_2=-\frac{1+\sqrt{5}}{4}+\frac{\sqrt{10-2\sqrt{5}}}{4}\mathrm{i}$$$$x_3=-\frac{1+\sqrt{5}}{4}-\frac{\sqrt{10-2\sqrt{5}}}{4}\mathrm{i}$$$$x_4=-\frac{1-\sqrt{5}}{4}-\frac{\sqrt{10+2\sqrt{5}}}{4}\mathrm{i}$$$$\Rightarrow$$$$x^4+x^3+x^2+x+1=$$$$=(x^2+\frac{1-\sqrt{5}}{2}x+1)(x^2+\frac{1+\sqrt{5}}{2}x+1)=$$$$=(x+\frac{1-\sqrt{5}}{4}-\frac{\sqrt{10+2\sqrt{5}}}{4}\mathrm{i})(x+\frac{1-\sqrt{5}}{4}+\frac{\sqrt{10+2\sqrt{5}}}{4}\mathrm{i})(x+\frac{1+\sqrt{5}}{4}-\frac{\sqrt{10-2\sqrt{5}}}{4}\mathrm{i})(x+\frac{1+\sqrt{5}}{4}+\frac{\sqrt{10-2\sqrt{5}}}{4}\mathrm{i})$$$$p\left({\mathrm{e}}^{\mathrm{i}\frac{\pi }{5}}\right)=1+\sqrt{5+2\sqrt{5}}\mathrm{i}$$$$p\left(\cos \frac{\pi }{5}\right)=\frac{67}{32}+\frac{19\sqrt{5}}{32}$$

Commented by mathmax by abdo last updated on 25/Aug/19

$$thankssir.$$

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