The triangle formed by the complex numbers z_ | vedclass

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(D) Let the vertices of the triangle be $A(z_1)$,$B(z_2)$,and $C(z_3)$,where $z_3 = -\omega z_1 - \omega^2 z_2$.For a triangle to be equilateral,the c

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equilateral

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(D) Let the vertices of the triangle be $A(z_1)$,$B(z_2)$,and $C(z_3)$,where $z_3 = -\omega z_1 - \omega^2 z_2$.For a triangle to be equilateral,the condition is $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$ or $z_1 + \omega z_2 + \omega^2 z_3 = 0$.Given $z_3 = -\omega z_1 - \omega^2 z_2$,we can rewrite this as $z_3 + \omega z_1 + \omega^2 z_2 = 0$.Since $1 + \omega + \omega^2 = 0$,we know $\omega^2 = -1 - \omega$.Substituting this into the condition for an equilateral triangle,we find that the vertices satisfy the property of an equilateral triangle.Thus,the triangle is equilateral.

The triangle formed by the complex numbers $z_1$,$z_2$,and $-\omega z_1 - \omega^2 z_2$ on the Argand plane is:

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