z 3 + i 2013 = 0 , z 3 = − i 2013 , z 3 = − i . z^3+i^{2013}=0, \qquad z^3=-i^{2013}, \qquad z^3=-i. z 3 + i 2013 = 0 , z 3 = − i 2013 , z 3 = − i . Prikažimo − i -i − i
− i = cos 3 π 2 + i sin 3 π 2 . -i=\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2}. − i = cos 2 3 π + i sin 2 3 π . z = − i 3 = cos 3 π 2 + 2 k π 3 + i sin 3 π 2 + 2 k π 3 , k = 0 , 1 , 2. z=\sqrt[3]{-i}=\cos\frac{\frac{3\pi}{2}+2k\pi}{3}+i\sin\frac{\frac{3\pi}{2}+2k\pi}{3},\quad k=0,1,2. z = 3 − i = cos 3 2 3 π + 2 k π + i sin 3 2 3 π + 2 k π , k = 0 , 1 , 2. Dakle, rješenja jednadžbe su sljedeći brojevi:
z 1 = cos π 2 + i sin π 2 = i , z 2 = cos 7 π 6 + i sin 7 π 6 = − 3 2 − 1 2 i , z 3 = cos 11 π 6 + i sin 11 π 6 = 3 2 − 1 2 i . \begin{aligned} z_1&=\cos\frac\pi2+i\sin\frac\pi2=i,\\ z_2&=\cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6}=-\frac{\sqrt3}{2}-\frac12i,\\ z_3&=\cos\frac{11\pi}{6}+i\sin\frac{11\pi}{6}=\frac{\sqrt3}{2}-\frac12i. \end{aligned} z 1 z 2 z 3 = cos 2 π + i sin 2 π = i , = cos 6 7 π + i sin 6 7 π = − 2 3 − 2 1 i , = cos 6 11 π + i sin 6 11 π = 2 3 − 2 1 i . Oni određuju vrhove jednakostraničnog trokuta
Z 1 ( 0 , 1 ) , Z 2 ( − 3 2 , − 1 2 ) , Z 3 ( 3 2 , − 1 2 ) . Z_1(0,1),\qquad Z_2\left(-\frac{\sqrt3}{2},-\frac12\right),\qquad Z_3\left(\frac{\sqrt3}{2},-\frac12\right). Z 1 ( 0 , 1 ) , Z 2 ( − 2 3 , − 2 1 ) , Z 3 ( 2 3 , − 2 1 ) .
Duljina stranice tog trokuta iznosi
a = ∣ Z 2 Z 3 ∣ = 2 ⋅ 3 2 = 3 . a=|Z_2Z_3|=2\cdot\frac{\sqrt3}{2}=\sqrt3. a = ∣ Z 2 Z 3 ∣ = 2 ⋅ 2 3 = 3 . Tada njegova površina iznosi
P = a 2 3 4 = 3 3 4 . P=\frac{a^2\sqrt3}{4}=\frac{3\sqrt3}{4}. P = 4 a 2 3 = 4 3 3 . 
