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Let a,b,c be distinct complex numbers such that a/1-b = b/1-c= c/1-a= k.
Find the value of k,where k is a constant.

Amit Saxena , 12 Years ago
Grade upto college level
anser 1 Answers
Radhika Batra
from  a/1-b = b / 1-c   we get                       b-b^2 = a - ac-----1

  similarly from the other 2 equations           a-a^2 = c - bc--------2
                                        
 we get                                                      c-c^2 = b - ab----------3

adding all three equations we get

a^2 + b^2 + c^2 = ab + bc + ca

as a,b,c are complex numbers the above equation is valid only
when a , b , c are the vertices of an equilateral triangle

now
-a/(b-1) = b/(1-c) = (b-a)/(b-c)   ---------rules of ratio and proportion
(b-a)/(b-c) = k

taking modulus on both sides , mod of (b-a) and (b-c) both are equal as they are the sides of equilateral triangle
so the values cancels out so the mod of k = 1

value of k = cos(pie/3) + i sin(pie/3)

Last Activity: 12 Years ago
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