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# if ABC is an equilateral triangle then if A and B are denoted by complex nos z1 and z2 then prove that C is denoted by z3 =z1 e^-i(pi\3)+z2 e^i(pi\3). Priyansh Bajaj AskiitiansExpert-IITD
30 Points
10 years ago

Dear Rachneet Kaur,

Solution:- By rotation theorem of complex no.'s, ((z3 - z1) / (z2 - z1)) = ( |z3 - z1| / |z2 - z1| )e^i(pi/3)

[Refer Rotation Theorem from http://www.askiitians.com/iit-jee-algebra/complex-numbers/rotation.aspx ]

Since ABC is an equilateral triangle, |z3 - z1| = |z2 - z1|

So, we get ((z3 - z1) / (z2 - z1)) = e^i(pi/3)

or, (z3 - z1) = e^i(pi/3) (z2 - z1)

or, z3 = z2 e^i(pi/3) + z1 (1-e^i(pi/3))

or, z3 = z2 e^i(pi/3) + z1 (1/2 - i (sqrt(3)/2))

or, z3 = z2 e^i(pi/3) + z1 e^-i(pi/3)

Hence Proved.

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