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Step-by-Step Solution
Step 1: Recall properties of the roots of the quadratic equation
Consider the quadratic equation
$Z^2 + aZ + b = 0$.
Its two roots are $Z_1$ and $Z_2$. From Vieta’s formulas:
$Z_1 + Z_2 = -a$
$Z_1 \, Z_2 = b$
Step 2: Understand the geometry of the equilateral triangle
We are told that the points corresponding to $0$, $Z_1$, and $Z_2$ form an equilateral triangle in the complex plane. One useful result for three complex numbers $z_1, z_2, z_3$ forming an equilateral triangle is:
$z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1.$
In our case, $z_1 = 0$, $z_2 = Z_1$, and $z_3 = Z_2$. Substituting into the above property gives:
$0^2 + Z_1^2 + Z_2^2 = 0 \times Z_1 + Z_1 \times Z_2 + Z_2 \times 0.$
This simplifies to:
$Z_1^2 + Z_2^2 = Z_1 Z_2.$
Step 3: Relate the condition to known sums and products
We already know $Z_1 \, Z_2 = b$, so:
$Z_1^2 + Z_2^2 = b.$
Next, realize that:
$(Z_1 + Z_2)^2 = Z_1^2 + 2Z_1 Z_2 + Z_2^2.$
Hence,
$Z_1^2 + Z_2^2 = (Z_1 + Z_2)^2 - 2Z_1 Z_2.$
Using $Z_1 + Z_2 = -a$ and $Z_1 Z_2 = b$, we get:
$Z_1^2 + Z_2^2 = (-a)^2 - 2b = a^2 - 2b.$
But from the equilateral triangle condition, we also have:
$Z_1^2 + Z_2^2 = b.$
Equating these two expressions for $Z_1^2 + Z_2^2$:
$a^2 - 2b = b.$
This gives:
$a^2 = 3b.$
Step 4: Conclusion
Therefore, the correct relationship is
$a^2 = 3b.$
