Prvo rješenje. Redom imamo:
z 3 = ( 1 + i ) 2 = 1 + 2 i − 1 = 2 i , 2 i = 2 ( cos π 2 + i sin π 2 ) ; z = 2 ( cos π 2 + i sin π 2 ) 3 , \begin{aligned}z^3&=(1+i)^2=1+2i-1=2i,\\2i&=2\left(\cos\frac\pi2+i\sin\frac\pi2\right);\\z&=\sqrt[3]{2\left(\cos\frac\pi2+i\sin\frac\pi2\right)},\end{aligned} z 3 2 i z = ( 1 + i ) 2 = 1 + 2 i − 1 = 2 i , = 2 ( cos 2 π + i sin 2 π ) ; = 3 2 ( cos 2 π + i sin 2 π ) , z k = 2 3 ( cos π 2 + 2 ( k − 1 ) π 3 + i sin π 2 + 2 ( k − 1 ) π 3 ) = 2 3 ( cos π + 4 ( k − 1 ) π 6 + i sin π + 4 ( k − 1 ) π 6 ) , k = 1 , 2 , 3 , \begin{aligned}z_k&=\sqrt[3]2\left(\cos\frac{\frac\pi2+2(k-1)\pi}3+i\sin\frac{\frac\pi2+2(k-1)\pi}3\right)\\&=\sqrt[3]2\left(\cos\frac{\pi+4(k-1)\pi}6+i\sin\frac{\pi+4(k-1)\pi}6\right),\ k=1,2,3,\end{aligned} z k = 3 2 ( cos 3 2 π + 2 ( k − 1 ) π + i sin 3 2 π + 2 ( k − 1 ) π ) = 3 2 ( cos 6 π + 4 ( k − 1 ) π + i sin 6 π + 4 ( k − 1 ) π ) , k = 1 , 2 , 3 , z 1 = 2 3 ( cos π 6 + i sin π 6 ) = 2 3 ( 3 2 + 1 2 i ) , z 2 = 2 3 ( cos 5 π 6 + i sin 5 π 6 ) = 2 3 ( − 3 2 + 1 2 i ) , z 3 = 2 3 ( cos 9 π 6 + i sin 9 π 6 ) = 2 3 ( cos 3 π 2 + i sin 3 π 2 ) = − 2 3 i . \begin{aligned}z_1&=\sqrt[3]2\left(\cos\frac\pi6+i\sin\frac\pi6\right)=\sqrt[3]2\left(\frac{\sqrt3}2+\frac12i\right),\\z_2&=\sqrt[3]2\left(\cos\frac{5\pi}6+i\sin\frac{5\pi}6\right)=\sqrt[3]2\left(-\frac{\sqrt3}2+\frac12i\right),\\z_3&=\sqrt[3]2\left(\cos\frac{9\pi}6+i\sin\frac{9\pi}6\right)=\sqrt[3]2\left(\cos\frac{3\pi}2+i\sin\frac{3\pi}2\right)=-\sqrt[3]2i.\end{aligned} z 1 z 2 z 3 = 3 2 ( cos 6 π + i sin 6 π ) = 3 2 ( 2 3 + 2 1 i ) , = 3 2 ( cos 6 5 π + i sin 6 5 π ) = 3 2 ( − 2 3 + 2 1 i ) , = 3 2 ( cos 6 9 π + i sin 6 9 π ) = 3 2 ( cos 2 3 π + i sin 2 3 π ) = − 3 2 i . Sada je
z 1 ⋅ z 2 ⋅ z 3 = − 2 i ( 3 2 + 1 2 i ) ( − 3 2 + 1 2 i ) = − 2 i ( − 1 4 − 3 4 ) = 2 i , z 1 2 ⋅ z 2 2 ⋅ z 3 2 = ( 2 i ) 2 = − 4. \begin{aligned}z_1\cdot z_2\cdot z_3&=-2i\left(\frac{\sqrt3}2+\frac12i\right)\left(-\frac{\sqrt3}2+\frac12i\right)=-2i\left(-\frac14-\frac34\right)=2i,\\z_1^2\cdot z_2^2\cdot z_3^2&=(2i)^2=-4.\end{aligned} z 1 ⋅ z 2 ⋅ z 3 z 1 2 ⋅ z 2 2 ⋅ z 3 2 = − 2 i ( 2 3 + 2 1 i ) ( − 2 3 + 2 1 i ) = − 2 i ( − 4 1 − 4 3 ) = 2 i , = ( 2 i ) 2 = − 4. Drugo rješenje. Zadatak se može riješiti i pomoću Vièteovih formula. Za jednadžbu trećeg stupnja a x 3 + b x 2 + c x + d = 0 ax^3+bx^2+cx+d=0 a x 3 + b x 2 + c x + d = 0 x 1 x_1 x 1 x 2 x_2 x 2 x 3 x_3 x 3 x 1 ⋅ x 2 ⋅ x 3 = − d a x_1\cdot x_2\cdot x_3=-\dfrac da x 1 ⋅ x 2 ⋅ x 3 = − a d
U danom slučaju imamo:
z 3 − ( 1 + i ) 2 = 0 , z 3 − 2 i = 0 , z 1 ⋅ z 2 ⋅ z 3 = 2 i , z 1 2 ⋅ z 2 2 ⋅ z 3 2 = ( 2 i ) 2 = − 4. \begin{aligned}z^3-(1+i)^2&=0,\\z^3-2i&=0,\\z_1\cdot z_2\cdot z_3&=2i,\\z_1^2\cdot z_2^2\cdot z_3^2&=(2i)^2=-4.\end{aligned} z 3 − ( 1 + i ) 2 z 3 − 2 i z 1 ⋅ z 2 ⋅ z 3 z 1 2 ⋅ z 2 2 ⋅ z 3 2 = 0 , = 0 , = 2 i , = ( 2 i ) 2 = − 4. 