Analyzing the Setup
Imagine you are standing on the complex plane, looking at the origin O(0) and two complex numbers Z1 and Z2. These three points form a perfectly symmetric equilateral triangle.
This geometric constraint implies rotational symmetry. If you rotate Z1 about the origin by 60∘, you land on Z2. This symmetry is the key to unlocking the relationship between the coefficients of the quadratic equation Z2+aZ+b=0.
The Algebraic Bridge
Vieta's Formulas
We are given the quadratic equation Z2+aZ+b=0. From the fundamental theorem of algebra, we know that the roots Z1 and Z2 are intimately connected to the coefficients a and b.
Specifically, the sum of the roots is given by:
Z1+Z2=−a
The product of the roots is given by:
Z1Z2=b
These two relations serve as our bridge between the abstract world of complex roots and the concrete world of coefficients.
The Equilateral Condition
A Powerful Tool
For any three complex numbers
Z1,Z2,Z3 to form an equilateral triangle, they must satisfy the following condition:
Z12+Z22+Z32=Z1Z2+Z2Z3+Z3Z1
In our specific case, one of the vertices is the origin, so we set Z3=0. When we substitute Z3=0 into this condition, the terms Z32, Z2Z3, and Z3Z1 vanish.
We are left with the simplified constraint:
Z12+Z22=Z1Z2
The Final Synthesis
We connect the geometric constraint
Z12+Z22=Z1Z2 with our algebraic relations using the identity:
Z12+Z22=(Z1+Z2)2−2Z1Z2
Substituting this into our constraint, we obtain:
(Z1+Z2)2−2Z1Z2=Z1Z2
Rearranging the terms leads to:
(Z1+Z2)2=3Z1Z2
Substituting the Vieta's relations
(−a)2=3(b), we arrive at the final condition:
a2=3b
A Moment of Celebration
We started with a geometric shape and, through the power of complex algebra, arrived at a clean, elegant condition. This is the beauty of mathematics, where the visual and the algebraic are woven into a single truth.
You have successfully navigated this classic JEE problem. Remember: every complex problem is just a simple truth waiting to be uncovered.