Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 30849 by Penguin last updated on 27/Feb/18

$$x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$$$$$$$$\underset{k=1}{\sum ^7}{\left[\mathrm{\Re }\left(x_k\right)\right]}^2=?$$$$x_k=k^{\mathrm{th}}\mathrm{root}\mathrm{of}\mathrm{the}\mathrm{equation}$$$$\mathrm{\Re }\left(x_k\right)=\mathrm{real}\mathrm{part}\mathrm{of}\mathrm{the}\mathrm{root}$$

Commented by prof Abdo imad last updated on 27/Feb/18

$$zrootforthisequation\iff z^8=1andz\ne o$$$$therootsofthisequationare{z}_k=e^{i\frac{k\pi }{4}}and$$$$k\in \left[[1,7]\right]\Rightarrow Re\left(z_k\right)=cos\left(\frac{k\pi }{4}\right)\Rightarrow$$$${\left(Re\left(z_k\right)\right)}^2={cos}^2\left(\frac{k\pi }{4}\right)and$$$$\sum _{k=1}^7{\left(Re\left(z_k\right)\right)}^2=\sum _{k=1}^7\frac{1+cos\left(\frac{k\pi }{2}\right)}{2}$$$$=\frac{7}{2}+\frac{1}{2}\sum _{k=1}^{7}cos\left(\frac{k\pi }{2}\right)but$$$$\sum _{k=1}^{7}cos\left(\frac{k\pi }{2}\right)=\sum _{k=0}^{7}cos\left(\frac{k\pi }{2}\right)-1$$$$=Re\left(\sum _{k=0}^7{e}^{i\frac{k\pi }{2}}\right)-1$$$$Re\left(\frac{1-{\left(e^{i\frac{\pi }{2}}\right)}^8}{1-e^{i\frac{\pi }{2}}}\right)-1=0-1=-1\Rightarrow$$$$\sum _{k=1}^7{\left(Re\left(z_k\right)\right)}^2=\frac{7}{2}-\frac{1}{2}=3.$$$$$$

Commented by Penguin last updated on 27/Feb/18

$$\sum _{k=1}^7{\left(Re\left(z_k\right)\right)}^2=\sum _{k=1}^7\frac{1+cos\left(\frac{k\pi }{2}\right)}{2}$$$$\mathrm{how}\mathrm{did}\mathrm{you}\mathrm{get}\mathrm{this}\mathrm{result}?$$

Commented by MJS last updated on 27/Feb/18

$$\mathrm{I}\mathrm{tried}\mathrm{to}\mathrm{solve}\mathrm{it},\mathrm{must}\mathrm{admit}\mathrm{that}$$$$\mathrm{this}\mathrm{might}\mathrm{not}\mathrm{always}\mathrm{be}\mathrm{possible}$$$$(\mathrm{but}\mathrm{I}\mathrm{always}\mathrm{love}\mathrm{to}\mathrm{try}...)$$$$x_1=-1$$$$x_2=-\mathrm{i}$$$$x_3=\mathrm{i}$$$$x_4=-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\mathrm{i}$$$$x_5=-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\mathrm{i}$$$$x_6=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\mathrm{i}$$$$x_7=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\mathrm{i}$$$$\underset{k=1}{\sum ^7}{\left[\mathrm{\Re }\left(x_k\right)\right]}^2=3$$

Commented by rahul 19 last updated on 27/Feb/18

$$yahh,shortmethods!$$

Commented by abdo imad last updated on 28/Feb/18

$$lookmyproofihaveusedtheformula{cos}^2\alpha =\frac{1+cos\left(2\alpha \right)}{2}.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com