Question

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

• A. circle with centre \((2, 0)\) and radius \(\sqrt{3}\)
• B. circle with centre \((0, 2)\) and radius \(\sqrt{3}\)
• C. circle with centre \((0, 0)\) and radius \(\sqrt{3}\)
• D. circle with centre \((0, -2)\) and radius \(\sqrt{3}\)

Hint:

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.

Answer 1

The Correct Option is B

Concept:
For \[ z=x+iy, \] we have \[ z+i = x+i(y+1), \qquad z-i = x+i(y-1) \] So the modulus equation becomes a relation between distances. ip

Step 1: Take modulus on numerator and denominator.
\[ \left|\frac{z+i}{z-i}\right|=\sqrt{3} \quad \Rightarrow \quad |z+i|=\sqrt{3}\,|z-i| \] Now, \[ |z+i|=\sqrt{x^2+(y+1)^2} \] and \[ |z-i|=\sqrt{x^2+(y-1)^2} \] So, \[ \sqrt{x^2+(y+1)^2}=\sqrt{3}\sqrt{x^2+(y-1)^2} \] ip

Step 2: Square both sides.
\[ x^2+(y+1)^2 = 3\left(x^2+(y-1)^2\right) \] \[ x^2+y^2+2y+1 = 3x^2+3y^2-6y+3 \] \[ 2x^2+2y^2-8y+2=0 \] \[ x^2+y^2-4y+1=0 \] ip

Step 3: Convert into standard circle form.
\[ x^2+y^2-4y+1=0 \] \[ x^2+(y^2-4y+4)=3 \] \[ x^2+(y-2)^2=3 \] So the circle has centre \[ (0,2) \] and radius \[ \sqrt{3} \] ip Hence, the correct answer is:
\[ \boxed{(B)\ \text{circle with centre }(0,2)\text{ and radius }\sqrt{3}} \]