Question:medium

If \(z = x + iy\) is a complex number, then the equation \(\left| \frac{z+i}{z-i} \right| = \sqrt{3}\) represents the

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Equations of the form \[ \left|\frac{z-z_1}{z-z_2}\right|=\text{constant} \] usually represent a circle or a line after simplifying the distance relation.
Updated On: May 14, 2026
  • circle with centre \((2, 0)\) and radius \(\sqrt{3}\)
  • circle with centre \((0, 2)\) and radius \(\sqrt{3}\)
  • circle with centre \((0, 0)\) and radius \(\sqrt{3}\)
  • circle with centre \((0, -2)\) and radius \(\sqrt{3}\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
The modulus equation represents a locus in the complex plane. We convert it to Cartesian coordinates by substituting \(z = x + iy\).
Step 2: Key Formula or Approach:
1. \(|x + iy| = \sqrt{x^2 + y^2}\).
2. Standard circle equation: \((x-h)^2 + (y-k)^2 = r^2\).
Step 3: Detailed Explanation:
The given equation is \(\frac{|z+i|}{|z-i|} = \sqrt{3}\), which implies \(|z+i|^2 = 3|z-i|^2\).
Substitute \(z = x+iy\):
\[ |x + i(y+1)|^2 = 3 |x + i(y-1)|^2 \] \[ x^2 + (y+1)^2 = 3[x^2 + (y-1)^2] \] \[ x^2 + y^2 + 2y + 1 = 3x^2 + 3y^2 - 6y + 3 \] Rearrange the terms into a standard form:
\[ 2x^2 + 2y^2 - 8y + 2 = 0 \] Divide by 2:
\[ x^2 + y^2 - 4y + 1 = 0 \] Complete the square for the \(y\) terms:
\[ x^2 + (y^2 - 4y + 4) = 4 - 1 \] \[ (x-0)^2 + (y-2)^2 = 3 \] This equation describes a circle with centre \((0, 2)\) and radius \(\sqrt{3}\).
Step 4: Final Answer:
The equation represents a circle with centre \((0, 2)\) and radius \(\sqrt{3}\).
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