Question:
If \(Q\) is real and \(z_1, z_2\) are connected by \(z_1^2 + z_2^2 + 2z_1z_2\cos \theta = 0\) then triangle with vertices \(0, z_1\) and \(z_2\) is
If \(Q\) is real and \(z_1, z_2\) are connected by \(z_1^2 + z_2^2 + 2z_1z_2\cos \theta = 0\) then triangle with vertices \(0, z_1\) and \(z_2\) is
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The ratio of complex numbers gives the relative magnitude and angle.
Updated On: Apr 23, 2026
- equilateral
- right-angled
- isosceles
- None of these
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Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Formula / Definition}
\[ \left(\frac{z_1}{z_2}\right)^2 + 2\cos\theta\left(\frac{z_1}{z_2}\right) + 1 = 0 \]
Step 2: Calculation / Simplification}
\(\frac{z_1}{z_2} = -\cos\theta \pm i\sin\theta\)
\(\left|\frac{z_1}{z_2}\right| = \sqrt{\cos^2\theta + \sin^2\theta} = 1 \Rightarrow |z_1| = |z_2|\)
Two sides from origin are equal \(\Rightarrow\) isosceles.
Step 3: Final Answer
\[ \text{isosceles} \]
\[ \left(\frac{z_1}{z_2}\right)^2 + 2\cos\theta\left(\frac{z_1}{z_2}\right) + 1 = 0 \]
Step 2: Calculation / Simplification}
\(\frac{z_1}{z_2} = -\cos\theta \pm i\sin\theta\)
\(\left|\frac{z_1}{z_2}\right| = \sqrt{\cos^2\theta + \sin^2\theta} = 1 \Rightarrow |z_1| = |z_2|\)
Two sides from origin are equal \(\Rightarrow\) isosceles.
Step 3: Final Answer
\[ \text{isosceles} \]
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