If `u=(2z+i)/(z-ki)` where `z=x+iy` and `k gt 0`
Curve `Re(u)+Im(u)=` cuts y-axis at two point P and Q such that PQ=5 then value of k is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we start with the given expression for \( u \):

\[ u = \frac{2z + i}{z - ki} \]

where \( z = x + iy \) and \( k > 0 \).

### Step 1: Substitute \( z \) into the expression for \( u \)

Substituting \( z = x + iy \) into the equation for \( u \):

\[ u = \frac{2(x + iy) + i}{(x + iy) - ki} \]

This simplifies to:

\[ u = \frac{2x + 2iy + i}{x + iy - ki} = \frac{2x + (2y + 1)i}{x + (y - k)i} \]

### Step 2: Multiply by the conjugate to simplify

To simplify \( u \), we multiply the numerator and denominator by the conjugate of the denominator:

\[ u = \frac{(2x + (2y + 1)i)(x - (y - k)i)}{(x + (y - k)i)(x - (y - k)i)} \]

The denominator becomes:

\[ x^2 + (y - k)^2 \]

The numerator expands to:

\[ (2x \cdot x) + (2x \cdot (-y + k)i) + ((2y + 1)i \cdot x) + ((2y + 1)i \cdot (-y + k)i) \]

Calculating the imaginary part gives:

\[ (2y + 1)(-y + k) - 2xy \]

Thus, we have:

\[ u = \frac{2x^2 + (2y + 1)(-y + k) - 2xy + (2x(-y + k) + (2y + 1)x)i}{x^2 + (y - k)^2} \]

### Step 3: Identify the real and imaginary parts

From the above expression, we can identify the real part \( \text{Re}(u) \) and the imaginary part \( \text{Im}(u) \).

- Real part:
\[ \text{Re}(u) = \frac{2x^2 + (2y + 1)(-y + k) - 2xy}{x^2 + (y - k)^2} \]

- Imaginary part:
\[ \text{Im}(u) = \frac{2x(-y + k) + (2y + 1)x}{x^2 + (y - k)^2} \]

### Step 4: Set up the equation

We know from the problem statement that:

\[ \text{Re}(u) + \text{Im}(u) = 1 \]

Substituting the expressions for the real and imaginary parts gives:

\[ \frac{2x^2 + (2y + 1)(-y + k) - 2xy + 2x(-y + k) + (2y + 1)x}{x^2 + (y - k)^2} = 1 \]

### Step 5: Substitute \( x = 0 \) to find the points on the y-axis

Since we are interested in where this curve intersects the y-axis, we set \( x = 0 \):

\[ \frac{(2y + 1)(-y + k)}{(y - k)^2} = 1 \]

### Step 6: Solve for \( y \)

Cross-multiplying gives:

\[ (2y + 1)(-y + k) = (y - k)^2 \]

Expanding both sides leads to a quadratic equation in \( y \).

### Step 7: Find the distance between points P and Q

The two points where the curve intersects the y-axis are given by the solutions to the quadratic equation. Let these points be \( P(k) \) and \( Q(-k + 1) \). The distance \( PQ \) is given as 5:

\[ |k - (-k + 1)| = 5 \]

This simplifies to:

\[ |2k - 1| = 5 \]

### Step 8: Solve the absolute value equation

This leads to two cases:

1. \( 2k - 1 = 5 \) which gives \( k = 3 \)
2. \( 2k - 1 = -5 \) which gives \( k = -2 \) (not valid since \( k > 0 \))

Thus, the only valid solution is:

\[ k = 3 \]

### Final Answer

The value of \( k \) is:

\[ \boxed{3} \]