SOLUTION: The points A, B represent complex number z1 and z2 respectively in the complex number plane, such that OAB is an equilateral tringle. O is the origin. Prove that {{{(z1)^2 + (z2)

Using trigonometric presentation of complex numbers (same as polar presentation),

I may assume that  = ,

where "r" is the modulus and  is the argument, i.e. the polar angle of the number .


Here I use the standard notation  =  for complex numbers.


From the condition,   =   with the same modulus "r" and the argument  rotated by 60°.


Since the formula   =  is uniform of the degree 2, the modules will cancel each other in both sides,

so I can forget about them and consider my complex numbers  and  as the unit vectors beginning at 0 (zero point).


So, I need to prove that 


     +  = .


You remember that when complex numbers are multiplied, their arguments added.


Therefore,   = ;   = .


In other words, the angle between unit vectors  and  is 120°.


So,  and  are the sides of the rhombus with the degree of 120° between them.


According to the "parallelogram rule" of adding complex numbers (and vectors),

the sum of the numbers  and  is the SHORT DIAGONAL of this rhombus,

i.e. the unit vector .


Thus I proved that   +  = .     (1)


From the other side,   =  = , 

i.e. the same complex number as (1).


Thus I proved that    +  = .