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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).

7 years ago

21 Points

Dear Rachneet,

z3-z2=(z2-z1)ei2π/3 (Rotating vector z2-z1 by external angle 2π/3 to get vector z3-z2)

z3=(1+ei2π/3)z2-z1ei2π/3

Now, 1+ei2π/3=-ei4π/3 (using 1+ω=-ω2) and -1=e-iπ

so 1+ei2π/3=-ei4π/3=ei(4π/3-π)=eiπ/3

also -ei2π/3=ei(2π/3-π)=e-iπ/3

z3=(1+ei2π/3)z2-z1ei2π/3 =>z3=z1e-iπ/3+z2eiπ/3

Hence proved.

Hope this helps.

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7 years ago
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