Question Number 19435 by Tinkutara last updated on 11/Aug/17
$$\mathrm{If}\:\alpha\:=\:\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{5}}\:+\:{i}\:\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{5}}\:,\:\mathrm{then}\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\alpha\:+\:\alpha^{\mathrm{2}} \:+\:\alpha^{\mathrm{3}} \:+\:\alpha^{\mathrm{4}} . \\ $$
Answered by ajfour last updated on 11/Aug/17
$$\underset{\mathrm{r}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}\alpha^{\mathrm{r}} =\frac{\alpha\left(\mathrm{1}−\alpha^{\mathrm{4}} \right)}{\mathrm{1}−\alpha}=\frac{\mathrm{e}^{\mathrm{i2}\pi/\mathrm{5}} −\mathrm{e}^{\mathrm{i2}\pi} }{\mathrm{1}−\mathrm{e}^{\mathrm{i2}\pi/\mathrm{5}} }=−\mathrm{1}\:. \\ $$$$ \\ $$
Commented by Tinkutara last updated on 11/Aug/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$