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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).
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. Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best Rachneet Kaur!!! Regards, Askiitians Experts Priyansh Bajaj
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.
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.
All the best Rachneet Kaur!!!
Regards,
Askiitians Experts
Priyansh Bajaj
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