#2_2015 Is it that complex??
Let z 1 , z 2 , z 3 are the roots of x 3 + 3 a x 2 + 3 b x + c = 0 in which a , b , c are complex numbers, corresponds to the points A,B,C on the complex plane. The triangle A B C is an equilateral triangle if,
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3 solutions
Good solution
I missed the answer I knew by pressing the wrong button !
Any way the lengths are equal, so the equation would be X^3=0.
:
x
3
+
3
a
x
2
+
3
a
2
x
+
a
3
=
(
x
+
a
)
3
=
0
So
b
=
a
2
is the answer.
The derivative has equal root
The roots of the equation z 1 , z 2 , z 3 represent the the vertices of of an triangle. The triangle is equilateral iff z 1 2 + z 2 2 + z 3 2 = z 1 z 2 + z 2 z 3 + z 3 z 1 . z 1 2 + z 2 2 + z 3 2 = ( z 1 + z 2 + z 3 ) 2 + 2 ( z 1 z 2 + z 2 z 3 + z 3 z 1 ) . Now, from the given equation we get that z 1 + z 2 + z 3 = − 3 a , z 1 z 2 + z 2 z 3 + z 3 z 1 = 3 b . By substituting all the values we get that 9 a 2 − 2 ( 3 b ) = 3 b therefore we get that a 2 = b
x 3 + 3 a x 2 + 3 b x + c = 0
Given the roots are z 1 , z 2 , z 3 ,
We know that z 1 + z 2 + z 3 = 1 − 3 a
Also, z 1 z 2 + z 2 z 3 + z 3 z 1 = 3 b
It is a theorem that if z 1 , z 2 , z 3 are vertices of an equilateral triangle , then
⇒ z 2 1 + z 2 2 + z 2 3 = z 1 z 2 + z 2 z 3 + z 3 z 1
⇒ ( z 1 + z 2 + z 3 ) 2 − 2 z 1 z 2 − 2 z 2 z 3 − 2 z 3 z 1 = z 1 z 2 + z 2 z 3 + z 3 z 1
⇒ ( z 1 + z 2 + z 3 ) 2 = 3 ( z 1 z 2 + z 2 z 3 + z 3 z 1 )
⇒ ( − 3 a ) 2 = 3 × 3 b
⇒ a 2 = b