if ABC is an equilateral triangle then if A and B are denoted by complex nos z1 and z2 then C is denoted by z3 =z1 e^-i(pi\3)+z2 e^i(pi\3).

Dear Rachneet,

z₃-z₂=(z₂-z₁)e^i2π/3 (Rotating vector z₂-z₁ by external angle 2π/3 to get vector z₃-z₂)

z₃=(1+e^i2π/3)z₂-z₁e^i2π/3 

Now, 1+e^i2π/3=-e^i4π/3 (using 1+ω=-ω²) and -1=e^-iπ 

so 1+e^i2π/3=-e^i4π/3=e^i(4π/3-π)=e^iπ/3 

also -e^i2π/3=e^i(2π/3-π)=e^-iπ/3

z₃=(1+e^i2π/3)z₂-z₁e^i2π/3 =>z₃=z₁e^-iπ/3+z₂e^iπ/3 

Hence proved.

Hope this helps.

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