Question

The locus of z such that  \(\frac{|z-i|}{|z+i|}\)= 2, where z = x+iy. is

• A. 3x² + 3y² +10y + 3
• B. 3x² - 3y² - 10y - 3 = 0
• C. 3x² + 3y² + 10y + 3 = 0
• D. x² + y² - 5y + 3 = 0

Answer 1

The Correct Option is C

To solve the problem, we need to find the locus of the complex number $z = x + iy$ that satisfies:

$\left| \frac{z - i}{z + i} \right| = 2$

1. Let $z = x + iy$:
Then, $z - i = x + i(y - 1)$, and
$z + i = x + i(y + 1)$

2. Express the modulus condition:
$\left| \frac{x + i(y - 1)}{x + i(y + 1)} \right| = 2$

3. Use modulus of a complex number:
$\left| \frac{a}{b} \right| = \frac{|a|}{|b|}$, so:
$\frac{\sqrt{x^2 + (y - 1)^2}}{\sqrt{x^2 + (y + 1)^2}} = 2$

4. Square both sides:
$\frac{x^2 + (y - 1)^2}{x^2 + (y + 1)^2} = 4$

5. Expand the squares:
Numerator: $x^2 + y^2 - 2y + 1$
Denominator: $x^2 + y^2 + 2y + 1$

6. Plug into equation:
$\frac{x^2 + y^2 - 2y + 1}{x^2 + y^2 + 2y + 1} = 4$

7. Cross-multiply:
$x^2 + y^2 - 2y + 1 = 4(x^2 + y^2 + 2y + 1)$

8. Expand both sides:
LHS: $x^2 + y^2 - 2y + 1$
RHS: $4x^2 + 4y^2 + 8y + 4$

9. Bring all terms to one side:
$x^2 + y^2 - 2y + 1 - 4x^2 - 4y^2 - 8y - 4 = 0$
Combine like terms: $-3x^2 - 3y^2 - 10y - 3 = 0$

10. Multiply both sides by -1:
$3x^2 + 3y^2 + 10y + 3 = 0$

Final Answer:
The locus of $z$ is given by the equation: ${3x^2 + 3y^2 + 10y + 3 = 0}$