If ` u = ( 2 z + 5 i)/( z - 3)` and | u| = 2 , then locus of z is a

Options

A. straight line
B. circle
C. parabola
D. none

Answer

To find the locus of the complex number \( z \) given the expression \( u = \frac{2z + 5i}{z - 3} \) and the condition \( |u| = 2 \), we can follow these steps:

### Step-by-Step Solution:

1. **Substitute \( z \) with \( x + iy \)**:
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. Then, we can rewrite \( u \):
\[ u = \frac{2(x + iy) + 5i}{(x + iy) - 3} = \frac{2x + 2iy + 5i}{x - 3 + iy} = \frac{2x + (2y + 5)i}{x - 3 + iy} \]

2. **Calculate the modulus of \( u \)**:
The modulus \( |u| \) is given by:
\[ |u| = \frac{|2x + (2y + 5)i|}{|x - 3 + iy|} \]
The modulus of the numerator is:
\[ |2x + (2y + 5)i| = \sqrt{(2x)^2 + (2y + 5)^2} = \sqrt{4x^2 + (2y + 5)^2} \]
The modulus of the denominator is:
\[ |x - 3 + iy| = \sqrt{(x - 3)^2 + y^2} \]

3. **Set up the equation using the given condition \( |u| = 2 \)**:
From the condition \( |u| = 2 \), we have:
\[ \frac{\sqrt{4x^2 + (2y + 5)^2}}{\sqrt{(x - 3)^2 + y^2}} = 2 \]

4. **Square both sides to eliminate the square roots**:
Squaring both sides gives:
\[ 4x^2 + (2y + 5)^2 = 4((x - 3)^2 + y^2) \]

5. **Expand both sides**:
The left-hand side expands to:
\[ 4x^2 + (4y^2 + 20y + 25) \]
The right-hand side expands to:
\[ 4(x^2 - 6x + 9 + y^2) = 4x^2 - 24x + 36 + 4y^2 \]

6. **Combine and simplify the equation**:
Setting both sides equal gives:
\[ 4x^2 + 4y^2 + 20y + 25 = 4x^2 - 24x + 36 + 4y^2 \]
Canceling \( 4x^2 \) and \( 4y^2 \) from both sides results in:
\[ 20y + 25 = -24x + 36 \]
Rearranging this gives:
\[ 24x + 20y = 11 \]

7. **Identify the locus**:
The equation \( 24x + 20y = 11 \) represents a straight line in the Cartesian plane.

### Conclusion:
The locus of \( z \) is a straight line given by the equation \( 24x + 20y = 11 \).