Question:
If $z_1,z_2$ are two roots of the equation
\[
z^2+az+b=0
\]
and on the Argand plane the points represented by $z_1,z_2$ and the origin form an equilateral triangle, then $a^2=$
If $z_1,z_2$ are two roots of the equation
\[
z^2+az+b=0
\]
and on the Argand plane the points represented by $z_1,z_2$ and the origin form an equilateral triangle, then $a^2=$
Show Hint
For roots forming an equilateral triangle with the origin, the angle between the roots is $60^\circ$.
Updated On: Jun 3, 2026
- $b$
- $2b$
- $3b$
- $4b$
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Concept
Use the geometric condition of an equilateral triangle in the complex plane.
Step 2: Meaning
Let the roots be \[ z_1=r(\cos\theta+i\sin\theta), \] \[ z_2=r(\cos(\theta+60^\circ)+i\sin(\theta+60^\circ)). \] Since the origin and the two roots form an equilateral triangle, \[ |z_1|=|z_2|=r. \]
Step 3: Analysis
By Vieta's formulas, \[ z_1+z_2=-a, \qquad z_1z_2=b. \] Now, \[ |z_1z_2|=r^2=b. \] Also, \[ |z_1+z_2| = 2r\cos30^\circ = \sqrt3\,r. \] Hence, \[ a^2=(z_1+z_2)^2=(\sqrt3\,r)^2=3r^2. \]
Step 4: Conclusion
Since $r^2=b$, \[ a^2=3b. \]
Final Answer: (C)
Use the geometric condition of an equilateral triangle in the complex plane.
Step 2: Meaning
Let the roots be \[ z_1=r(\cos\theta+i\sin\theta), \] \[ z_2=r(\cos(\theta+60^\circ)+i\sin(\theta+60^\circ)). \] Since the origin and the two roots form an equilateral triangle, \[ |z_1|=|z_2|=r. \]
Step 3: Analysis
By Vieta's formulas, \[ z_1+z_2=-a, \qquad z_1z_2=b. \] Now, \[ |z_1z_2|=r^2=b. \] Also, \[ |z_1+z_2| = 2r\cos30^\circ = \sqrt3\,r. \] Hence, \[ a^2=(z_1+z_2)^2=(\sqrt3\,r)^2=3r^2. \]
Step 4: Conclusion
Since $r^2=b$, \[ a^2=3b. \]
Final Answer: (C)
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