#2_2015 Is it that complex??

Algebra Level 4

Let z 1 , z 2 , z 3 { z }_{ 1 },{ z }_{ 2 },{ z }_{ 3 } are the roots of x 3 + 3 a x 2 + 3 b x + c = 0 { x }^{ 3 }+3a{ x }^{ 2 }+3bx+c=0 in which a , b , c a,b,c are complex numbers, corresponds to the points A,B,C on the complex plane. The triangle A B C ABC is an equilateral triangle if,

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a 2 > 3 b { a }^{ 2 }>3b a 2 < 3 b { a }^{ 2 }<3b a 2 = b { a }^{ 2 }=b a 2 = 3 b { a }^{ 2 }=3b

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3 solutions

Anandhu Raj
Jan 14, 2015

x 3 + 3 a x 2 + 3 b x + c = 0 { x }^{ 3 }+3a{ x }^{ 2 }+3bx+c=0

Given the roots are z 1 , z 2 , z 3 { z }_{ 1 },{ z }_{ 2 },{ z }_{ 3 } ,

We know that z 1 + z 2 + z 3 { z }_{ 1 }+{ z }_{ 2 }{ +z }_{ 3 } = 3 a 1 \frac { -3a }{ 1 }

Also, z 1 z 2 + z 2 z 3 + z 3 z 1 = 3 b { z }_{ 1 }{ z }_{ 2 }+{ z }_{ 2 }{ z }_{ 3 }+{ z }_{ 3 }{ z }_{ 1 }=3b

It is a theorem that if z 1 , z 2 , z 3 { z }_{ 1 },{ z }_{ 2 },{ z }_{ 3 } are vertices of an equilateral triangle , then

\Rightarrow z 2 1 { z^{ 2 } }_{ 1 } + z 2 2 { z^{ 2 } }_{ 2 } + z 2 3 { z^{ 2 } }_{ 3 } = z 1 z 2 + z 2 z 3 + z 3 z 1 { z }_{ 1 }{ z }_{ 2 }+{ z }_{ 2 }{ z }_{ 3 }+{ z }_{ 3 }{ z }_{ 1 }

\Rightarrow ( 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 }-2{ z }_{ 1 }{ z }_{ 2 }-2{ z }_{ 2 }{ z }_{ 3 }-2{ z }_{ 3 }{ z }_{ 1 }\quad =\quad { z }_{ 1 }{ z }_{ 2 }+{ z }_{ 2 }{ z }_{ 3 }+{ z }_{ 3 }{ z }_{ 1 }

\Rightarrow ( z 1 + z 2 + z 3 ) 2 = 3 ( z 1 z 2 + z 2 z 3 + z 3 z 1 ) ({ z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 })^{ 2 }\quad =\quad 3({ z }_{ 1 }{ z }_{ 2 }+{ z }_{ 2 }{ z }_{ 3 }+{ z }_{ 3 }{ z }_{ 1 })

\Rightarrow ( 3 a ) 2 = 3 × 3 b (-3a)^{ 2 }\quad =\quad 3\times 3b

a 2 = b \Rightarrow { a }^{ 2 }\quad =\quad b

Good solution

U Z - 6 years, 3 months ago

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@megh choksi Thanks :)

Anandhu Raj - 6 years, 2 months ago

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 { x }^{ 3 }+3a{ x }^{ 2 }+3a^2x+a^3=(x+a)^3=0
So b = a 2 b=a^2 is the answer.

Niranjan Khanderia - 4 years, 11 months ago
Vagish Jha
Sep 15, 2015

Use Stegner ellipse concept

The derivative has equal root

Vagish Jha - 5 years, 9 months ago
Mihir Chakravarti
Jan 14, 2015

The roots of the equation z 1 , z 2 , z 3 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_{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 ) 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 z_{1} + z_{2} + z_{3} = -3a, z_{1}z_{2} + z_{2}z_{3} + z_{3}z_{1} = 3b . By substituting all the values we get that 9 a 2 2 ( 3 b ) = 3 b 9a^{2} - 2(3b) = 3b therefore we get that a 2 = b \boxed{a^{2} = b}

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