Question

Daca ε este o solutie a ecuatiei x^2+x+1=0 sa se calculeze 1 + ε + ε^2 + ... + ε^2013

Answer 1

x²+x+1=0

Δ=1-4=-3

√Δ=√-3=i√3

$x_1=\frac{-1+i\sqrt{3} }{2} \\\\x_2=\frac{-1-i\sqrt{3} }{2} \\\\\\\varepsilon=\frac{-1+i\sqrt{3} }{2}$

$\varepsilon^2=\frac{-2-2i\sqrt{3} }{4} =\frac{-2(1+i\sqrt{3}) }{4} =-\frac{1+i\sqrt{3} }{2}$

$\varepsilon^3=\frac{1+3}{4} =1$

$\varepsilon^3=element\ neutru \implies \varepsilon^4=\varepsilon\\\\\varepsilon^5=\varepsilon^2\\\\\varepsilon^6=\varepsilo1\\\\...\\\varepsilon^2^0^1^3=1$

2013:3=671 rest 0

S=1+ε+ε²+1+ε+ε²+...+1+ε+ε²+1

Avem 2014 termeni

2014:3=671 rest 1

Deci avem 671 grupe de 1+ε+ε²  plus restul 1, adica termenul 1

S=671(1+ε+ε²)+1

Calculam 1+ε+ε²

$1+\varepsilon+\varepsilon^2=1+\frac{-1+i\sqrt{3} }{2} -\frac{1+i\sqrt{3} }{2} =1+\frac{-2}{2} =1-1=0$

S=671×0+1=1