If `|(z+i)/(z-i)|=sqrt(3)`, then z lies on a circle whose radius, is

Options

A. `2/sqrt(21)`
B. `1/sqrt(21)`
C. `sqrt(3)`
D. `sqrt(21)`

Explanation

To solve the problem \( \left| \frac{z+i}{z-i} \right| = \sqrt{3} \), we will follow these steps:

### Step 1: Rewrite the modulus equation
We start with the given equation:
\[ \left| \frac{z+i}{z-i} \right| = \sqrt{3} \]
This can be rewritten using the property of moduli:
\[ \frac{|z+i|}{|z-i|} = \sqrt{3} \]

### Step 2: Cross-multiply
Cross-multiplying gives us:
\[ |z+i| = \sqrt{3} |z-i| \]

### Step 3: Express \( z \) in terms of \( x \) and \( y \)
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. Then we can express the moduli:
\[ |z+i| = |x + i(y+1)| = \sqrt{x^2 + (y+1)^2} \]
\[ |z-i| = |x + i(y-1)| = \sqrt{x^2 + (y-1)^2} \]

### Step 4: Substitute moduli into the equation
Substituting the expressions for the moduli into our equation gives:
\[ \sqrt{x^2 + (y+1)^2} = \sqrt{3} \sqrt{x^2 + (y-1)^2} \]

### Step 5: Square both sides
Squaring both sides to eliminate the square roots results in:
\[ x^2 + (y+1)^2 = 3(x^2 + (y-1)^2) \]

### Step 6: Expand both sides
Expanding both sides:
\[ x^2 + (y^2 + 2y + 1) = 3(x^2 + (y^2 - 2y + 1)) \]
This simplifies to:
\[ x^2 + y^2 + 2y + 1 = 3x^2 + 3y^2 - 6y + 3 \]

### Step 7: Rearrange the equation
Rearranging gives:
\[ x^2 + y^2 + 2y + 1 - 3x^2 - 3y^2 + 6y - 3 = 0 \]
\[ -2x^2 - 2y^2 + 8y - 2 = 0 \]

### Step 8: Divide by -2
Dividing the entire equation by -2:
\[ x^2 + y^2 - 4y + 1 = 0 \]

### Step 9: Complete the square
Completing the square for the \( y \) terms:
\[ x^2 + (y^2 - 4y + 4) - 4 + 1 = 0 \]
\[ x^2 + (y-2)^2 - 3 = 0 \]
This simplifies to:
\[ x^2 + (y-2)^2 = 3 \]

### Step 10: Identify the circle
This equation represents a circle centered at \( (0, 2) \) with a radius of \( \sqrt{3} \).

### Conclusion
Thus, the radius of the circle on which \( z \) lies is:
\[ \text{Radius} = \sqrt{3} \]

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