De Moivre's formula

There are two distinct complex numbers z such that z3 is equal to 1 and z is not equal 1.

Calculate the sum of these two numbers.

Correct result:

S =  -1

Solution:

z3=1 1=1(cos0+isin0) zk=13(cos(2πk3)+isin(2πk3)) z1=1 z2=cos(120)+isin(120)=0.5+0.86602540378444i z3=cos(240)+isin(240)=0.50.86602540378444i S=z2+z3=1z^3 = 1 \ \\ 1 = 1(\cos 0 + i \sin 0 ) \ \\ z_k = \sqrt[3]{1}( \cos( \dfrac{2\pi k }{3} ) + i \sin( \dfrac{2\pi k }{3} ) ) \ \\ z_1 = 1 \ \\ z_2 = \cos( 120 ^\circ ) + i \sin( 120 ^\circ ) = -0.5 + 0.86602540378444 i \ \\ z_3 = \cos( 240 ^\circ ) + i \sin( 240 ^\circ ) = -0.5 - 0.86602540378444 i \ \\ S = z_2 + z_3 = -1



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