For a complex number Z, if the argument of `(Z-a)(barZ-b)` is `(pi)/(4)` or `(-3pi)/(4)` (where a, b are two real numbers), then the value of ab such that the locus of Z represents a circle with centre `(3)/(2)+(i)/(2)` is (where, `i^(2)=-1`)

Options

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

Answer

To solve the problem, we need to analyze the given complex number \( Z \) and the conditions provided. Let's break down the solution step by step.

### Step 1: Understand the given expression
We are given that the argument of \( (Z - a)(\overline{Z} - b) \) is either \( \frac{\pi}{4} \) or \( -\frac{3\pi}{4} \). This means that the expression \( (Z - a)(\overline{Z} - b) \) lies on a line at an angle of \( \frac{\pi}{4} \) or \( -\frac{3\pi}{4} \) with respect to the positive x-axis.

### Step 2: Express \( Z \) in terms of its components
Let \( Z = x + iy \), where \( x \) and \( y \) are the real and imaginary parts of \( Z \), respectively. The conjugate of \( Z \) is \( \overline{Z} = x - iy \).

### Step 3: Substitute into the expression
Now, substituting \( Z \) and \( \overline{Z} \) into the expression:
\[ (Z - a)(\overline{Z} - b) = (x + iy - a)(x - iy - b) \]
This simplifies to:
\[ = ((x - a) + iy)((x - b) - iy) \]
Expanding this:
\[ = (x - a)(x - b) + i(y(x - b) - y(x - a)) + y^2 \]
This can be further simplified to:
\[ = (x - a)(x - b) + y^2 + i(y(x - b) - y(x - a)) \]

### Step 4: Find the argument condition
The argument condition states that:
\[ \frac{y(x - b) - y(x - a)}{(x - a)(x - b) + y^2} = \tan\left(\frac{\pi}{4}\right) \text{ or } \tan\left(-\frac{3\pi}{4}\right) \]
Both conditions imply that:
\[ y(x - b) - y(x - a) = \pm((x - a)(x - b) + y^2) \]

### Step 5: Rearranging the equation
We can rearrange this to isolate terms involving \( x \) and \( y \):
\[ y(x - b) - y(x - a) = 0 \implies y(a - b) = 0 \]
This means either \( y = 0 \) or \( a = b \).

### Step 6: Locus of \( Z \)
We know that the locus of \( Z \) represents a circle with center \( \frac{3}{2} + \frac{i}{2} \). The general equation of a circle is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \( (h, k) \) is the center. Here, \( h = \frac{3}{2} \) and \( k = \frac{1}{2} \).

### Step 7: Equating the center
From our earlier analysis, the center of the circle can be found as:
\[ \left(\frac{a + b}{2}, \frac{a - b}{2}\right) = \left(\frac{3}{2}, \frac{1}{2}\right) \]
This gives us two equations:
1. \( \frac{a + b}{2} = \frac{3}{2} \) implies \( a + b = 3 \)
2. \( \frac{a - b}{2} = \frac{1}{2} \) implies \( a - b = 1 \)

### Step 8: Solve the system of equations
Now we can solve the system of equations:
1. \( a + b = 3 \)
2. \( a - b = 1 \)

Adding these two equations:
\[ 2a = 4 \implies a = 2 \]
Substituting \( a = 2 \) into \( a + b = 3 \):
\[ 2 + b = 3 \implies b = 1 \]

### Step 9: Find the product \( ab \)
Now, we can find \( ab \):
\[ ab = 2 \times 1 = 2 \]

### Final Answer
Thus, the value of \( ab \) is \( \boxed{2} \).