Nultočke trinoma x 2 + x + 1 x^2+x+1 x 2 + x + 1 − 1 2 ± 3 2 i -\dfrac12\pm\dfrac{\sqrt3}{2}i − 2 1 ± 2 3 i n ∈ N n\in\mathbb N n ∈ N
( − 1 2 ± 3 2 i + 1 ) n − ( − 1 2 ± 3 2 i ) n − 1 = ( 1 2 ± 3 2 i ) n − ( − 1 2 ± 3 2 i ) n − 1 = 0. \begin{aligned}&\left(-\frac12\pm\frac{\sqrt3}{2}i+1\right)^n-\left(-\frac12\pm\frac{\sqrt3}{2}i\right)^n-1\\&=\left(\frac12\pm\frac{\sqrt3}{2}i\right)^n-\left(-\frac12\pm\frac{\sqrt3}{2}i\right)^n-1=0.\end{aligned} ( − 2 1 ± 2 3 i + 1 ) n − ( − 2 1 ± 2 3 i ) n − 1 = ( 2 1 ± 2 3 i ) n − ( − 2 1 ± 2 3 i ) n − 1 = 0. Kako je − 1 2 ± 3 2 i = cos 2 π 3 ± i sin 2 π 3 -\dfrac12\pm\dfrac{\sqrt3}{2}i=\cos\dfrac{2\pi}{3}\pm i\sin\dfrac{2\pi}{3} − 2 1 ± 2 3 i = cos 3 2 π ± i sin 3 2 π 1 2 ± 3 2 i = cos π 3 ± i sin π 3 \dfrac12\pm\dfrac{\sqrt3}{2}i=\cos\dfrac\pi3\pm i\sin\dfrac\pi3 2 1 ± 2 3 i = cos 3 π ± i sin 3 π − 1 2 ± 3 2 i -\dfrac12\pm\dfrac{\sqrt3}{2}i − 2 1 ± 2 3 i 1 2 ± 3 2 i \dfrac12\pm\dfrac{\sqrt3}{2}i 2 1 ± 2 3 i ( x 3 − 1 ) (x^3-1) ( x 3 − 1 ) ( x 3 + 1 ) (x^3+1) ( x 3 + 1 )
( − 1 2 ± 3 2 i ) 3 = 1 i ( 1 2 ± 3 2 i ) 3 = − 1. \left(-\frac12\pm\frac{\sqrt3}{2}i\right)^3=1\quad\text{i}\quad\left(\frac12\pm\frac{\sqrt3}{2}i\right)^3=-1. ( − 2 1 ± 2 3 i ) 3 = 1 i ( 2 1 ± 2 3 i ) 3 = − 1. Zato je dovoljno promatrati slučajeve n = 3 k n=3k n = 3 k n = 3 k + 1 n=3k+1 n = 3 k + 1 n = 3 k + 2 n=3k+2 n = 3 k + 2
1 ∘ 1^\circ 1 ∘ n = 3 k n=3k n = 3 k
( 1 2 ± 3 2 i ) 3 k − ( − 1 2 ± 3 2 i ) 3 k − 1 = ( − 1 ) k − 1 k − 1 = ( − 1 ) k ≠ 0. \left(\frac12\pm\frac{\sqrt3}{2}i\right)^{3k}-\left(-\frac12\pm\frac{\sqrt3}{2}i\right)^{3k}-1=(-1)^k-1^k-1=(-1)^k\neq0. ( 2 1 ± 2 3 i ) 3 k − ( − 2 1 ± 2 3 i ) 3 k − 1 = ( − 1 ) k − 1 k − 1 = ( − 1 ) k = 0. 2 ∘ 2^\circ 2 ∘ n = 3 k + 1 n=3k+1 n = 3 k + 1
( 1 2 ± 3 2 i ) 3 k + 1 − ( − 1 2 ± 3 2 i ) 3 k + 1 − 1 = ( − 1 ) k ( 1 2 ± 3 2 i ) − ( − 1 2 ± 3 2 i ) − 1 = … \begin{aligned}&\left(\frac12\pm\frac{\sqrt3}{2}i\right)^{3k+1}-\left(-\frac12\pm\frac{\sqrt3}{2}i\right)^{3k+1}-1\\&=(-1)^k\left(\frac12\pm\frac{\sqrt3}{2}i\right)-\left(-\frac12\pm\frac{\sqrt3}{2}i\right)-1=\ldots\end{aligned} ( 2 1 ± 2 3 i ) 3 k + 1 − ( − 2 1 ± 2 3 i ) 3 k + 1 − 1 = ( − 1 ) k ( 2 1 ± 2 3 i ) − ( − 2 1 ± 2 3 i ) − 1 = … Za neparno k k k
… = ( − 1 ) ( 1 2 ± 3 2 i ) − ( − 1 2 ± 3 2 i ) − 1 = − 1 ∓ 3 ≠ 0 , \ldots=(-1)\left(\frac12\pm\frac{\sqrt3}{2}i\right)-\left(-\frac12\pm\frac{\sqrt3}{2}i\right)-1=-1\mp\sqrt3\neq0, … = ( − 1 ) ( 2 1 ± 2 3 i ) − ( − 2 1 ± 2 3 i ) − 1 = − 1 ∓ 3 = 0 , a za parno k k k
… = ( 1 2 ± 3 2 i ) − ( − 1 2 ± 3 2 i ) − 1 = 0. \ldots=\left(\frac12\pm\frac{\sqrt3}{2}i\right)-\left(-\frac12\pm\frac{\sqrt3}{2}i\right)-1=0. … = ( 2 1 ± 2 3 i ) − ( − 2 1 ± 2 3 i ) − 1 = 0. 3 ∘ 3^\circ 3 ∘ n = 3 k + 2 n=3k+2 n = 3 k + 2
( 1 2 ± 3 2 i ) 3 k + 2 − ( − 1 2 ± 3 2 i ) 3 k + 2 − 1 = ( − 1 ) k ( − 1 2 ± 3 2 i ) − ( − 1 2 ∓ 3 2 i ) − 1 = … \begin{aligned}&\left(\frac12\pm\frac{\sqrt3}{2}i\right)^{3k+2}-\left(-\frac12\pm\frac{\sqrt3}{2}i\right)^{3k+2}-1\\&=(-1)^k\left(-\frac12\pm\frac{\sqrt3}{2}i\right)-\left(-\frac12\mp\frac{\sqrt3}{2}i\right)-1=\ldots\end{aligned} ( 2 1 ± 2 3 i ) 3 k + 2 − ( − 2 1 ± 2 3 i ) 3 k + 2 − 1 = ( − 1 ) k ( − 2 1 ± 2 3 i ) − ( − 2 1 ∓ 2 3 i ) − 1 = … Za neparno k k k
… = − ( − 1 2 ± 3 2 i ) − ( − 1 2 ∓ 3 2 i ) − 1 = 0 , \ldots=-\left(-\frac12\pm\frac{\sqrt3}{2}i\right)-\left(-\frac12\mp\frac{\sqrt3}{2}i\right)-1=0, … = − ( − 2 1 ± 2 3 i ) − ( − 2 1 ∓ 2 3 i ) − 1 = 0 , a za parno k k k
… = ( − 1 2 ± 3 2 i ) − ( − 1 2 ∓ 3 2 i ) − 1 = − 1 ± 3 i ≠ 0. \ldots=\left(-\frac12\pm\frac{\sqrt3}{2}i\right)-\left(-\frac12\mp\frac{\sqrt3}{2}i\right)-1=-1\pm\sqrt3i\neq0. … = ( − 2 1 ± 2 3 i ) − ( − 2 1 ∓ 2 3 i ) − 1 = − 1 ± 3 i = 0. Dakle, zadovoljavaju brojevi oblika n = 3 ⋅ 2 l + 1 n=3\cdot2l+1 n = 3 ⋅ 2 l + 1 n = 3 ( 2 l + 1 ) + 2 n=3(2l+1)+2 n = 3 ( 2 l + 1 ) + 2 n = 6 l + 1 n=6l+1 n = 6 l + 1 n = 6 l + 5 n=6l+5 n = 6 l + 5 l ∈ N 0 l\in\mathbb N_0 l ∈ N 0
Napomena. Ovo se može malo elegantnije zapisati pomoću trigonometrijskog ili eksponencijalnog oblika kompleksnog broja.
