If `z(bar(z+alpha))+barz(z+alpha)=0`, where `alpha` is a complex constant, then z is represented by a point on

Options

A. a circle
B. a straight line
C. a parabola
D. none of these

Answer

To solve the equation \( z(\overline{z + \alpha}) + \overline{z}(z + \alpha) = 0 \), where \( \alpha \) is a complex constant, we will follow these steps:

### Step 1: Rewrite the equation
Start with the given equation:
\[ z(\overline{z + \alpha}) + \overline{z}(z + \alpha) = 0 \]
Using the property of conjugates, we can rewrite \( \overline{z + \alpha} \) as \( \overline{z} + \overline{\alpha} \). Thus, the equation becomes:
\[ z(\overline{z} + \overline{\alpha}) + \overline{z}(z + \alpha) = 0 \]

### Step 2: Expand the equation
Now, expand both terms:
\[ z\overline{z} + z\overline{\alpha} + \overline{z}z + \overline{z}\alpha = 0 \]
Notice that \( z\overline{z} = |z|^2 \), so we can combine like terms:
\[ |z|^2 + |z|^2 + z\overline{\alpha} + \overline{z}\alpha = 0 \]
This simplifies to:
\[ 2|z|^2 + z\overline{\alpha} + \overline{z}\alpha = 0 \]

### Step 3: Rearranging the equation
Rearranging gives:
\[ 2|z|^2 + z\overline{\alpha} + \overline{z}\alpha = 0 \]
This can be rewritten as:
\[ 2|z|^2 = - (z\overline{\alpha} + \overline{z}\alpha) \]

### Step 4: Substitute \( z = x + iy \)
Let \( z = x + iy \) and \( \alpha = a + ib \) where \( a, b \) are real numbers. Then:
\[ \overline{z} = x - iy \]
Substituting these into the equation gives:
\[ 2(x^2 + y^2) = -((x + iy)(a - ib) + (x - iy)(a + ib)) \]

### Step 5: Expand and simplify
Now, expand the right-hand side:
\[ = -[(ax + by) + i(ay - bx) + (ax + by) - i(ay - bx)] \]
The imaginary parts cancel out, resulting in:
\[ = -2(ax + by) \]
Thus, we have:
\[ 2(x^2 + y^2) = -2(ax + by) \]
Dividing through by 2 gives:
\[ x^2 + y^2 + ax + by = 0 \]

### Step 6: Complete the square
To rewrite this in standard form, complete the square:
\[ x^2 + ax + y^2 + by = 0 \]
Completing the square for \( x \) and \( y \):
\[ \left(x + \frac{a}{2}\right)^2 - \frac{a^2}{4} + \left(y + \frac{b}{2}\right)^2 - \frac{b^2}{4} = 0 \]
Rearranging gives:
\[ \left(x + \frac{a}{2}\right)^2 + \left(y + \frac{b}{2}\right)^2 = \frac{a^2 + b^2}{4} \]

### Conclusion
This is the equation of a circle with center at \((- \frac{a}{2}, - \frac{b}{2})\) and radius \(\sqrt{\frac{a^2 + b^2}{4}}\).

Thus, the solution is that \( z \) is represented by a point on a **circle**.