The region of the complex plane for which `|(z-a)/(z+veca)|=1,(Re(a) != 0)` is

Options

A. x-axis
B. y-axis
C. Straight line x=a
D. The straight line y=a

Answer

To solve the problem, we need to analyze the equation given in the complex plane:

\[ \left| \frac{z - a}{z + \bar{a}} \right| = 1 \]

where \( z \) is a complex number and \( a \) is a complex number with \( \text{Re}(a) \neq 0 \).

### Step 1: Rewrite the complex numbers
Let \( z = x + iy \) and \( a = b + ic \), where \( b = \text{Re}(a) \) and \( c = \text{Im}(a) \). The conjugate of \( a \) is \( \bar{a} = b - ic \).

### Step 2: Substitute into the equation
Substituting \( z \) and \( a \) into the equation gives:

\[ \left| \frac{(x + iy) - (b + ic)}{(x + iy) + (b - ic)} \right| = 1 \]

This simplifies to:

\[ \left| \frac{(x - b) + i(y - c)}{(x + b) + i(y - c)} \right| = 1 \]

### Step 3: Use the property of modulus
The modulus of a complex number \( \frac{u + iv}{x + iy} \) is given by:

\[ \left| \frac{u + iv}{x + iy} \right| = \frac{\sqrt{u^2 + v^2}}{\sqrt{x^2 + y^2}} \]

Thus, we can write:

\[ \sqrt{(x - b)^2 + (y - c)^2} = \sqrt{(x + b)^2 + (y - c)^2} \]

### Step 4: Square both sides
Squaring both sides gives:

\[ (x - b)^2 + (y - c)^2 = (x + b)^2 + (y - c)^2 \]

### Step 5: Expand both sides
Expanding both sides results in:

\[ (x^2 - 2bx + b^2 + y^2 - 2cy + c^2) = (x^2 + 2bx + b^2 + y^2 - 2cy + c^2) \]

### Step 6: Simplify the equation
Canceling out common terms \( x^2, y^2, b^2, c^2, \) and \( -2cy \) from both sides gives:

\[ -2bx = 2bx \]

### Step 7: Rearranging the equation
Rearranging this leads to:

\[ -4bx = 0 \]

### Step 8: Solving for \( x \)
Since \( b \neq 0 \) (as given \( \text{Re}(a) \neq 0 \)), we can conclude:

\[ x = 0 \]

### Step 9: Conclusion
The solution \( x = 0 \) indicates that the locus of points \( z \) lies on the imaginary axis in the complex plane.

### Final Answer
The region of the complex plane for which \( \left| \frac{z - a}{z + \bar{a}} \right| = 1 \) is the imaginary axis.

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