If ` z^(2) + z | z| + | z|^(2) = 0 ` , then the locus of z is

Options

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

Answer

To solve the equation \( z^2 + z |z| + |z|^2 = 0 \) and find the locus of \( z \), we can follow these steps:

### Step 1: Represent \( z \) in terms of its real and imaginary parts
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers.

### Step 2: Calculate \( z^2 \)
Using the formula for squaring a complex number:
\[ z^2 = (x + iy)^2 = x^2 + 2xyi - y^2 = (x^2 - y^2) + 2xyi \]

### Step 3: Calculate \( |z| \)
The modulus of \( z \) is given by:
\[ |z| = \sqrt{x^2 + y^2} \]

### Step 4: Substitute into the equation
Now substitute \( z^2 \), \( z|z| \), and \( |z|^2 \) into the original equation:
\[ (x^2 - y^2) + 2xyi + (x + iy)\sqrt{x^2 + y^2} + (x^2 + y^2) = 0 \]

### Step 5: Simplify the equation
We can separate the real and imaginary parts:
- Real part: \( x^2 - y^2 + x\sqrt{x^2 + y^2} + x^2 + y^2 = 0 \)
- Imaginary part: \( 2xy + y\sqrt{x^2 + y^2} = 0 \)

### Step 6: Solve the imaginary part
From the imaginary part:
\[ y(2x + \sqrt{x^2 + y^2}) = 0 \]
This gives us two cases:
1. \( y = 0 \)
2. \( 2x + \sqrt{x^2 + y^2} = 0 \)

#### Case 1: \( y = 0 \)
If \( y = 0 \), then \( z = x \) and we substitute into the real part:
\[ x^2 + x^2 = 0 \implies 2x^2 = 0 \implies x = 0 \]
Thus, \( z = 0 \).

#### Case 2: \( 2x + \sqrt{x^2 + y^2} = 0 \)
Squaring both sides:
\[ (2x)^2 = x^2 + y^2 \implies 4x^2 = x^2 + y^2 \implies 3x^2 = y^2 \implies y = \pm \sqrt{3}x \]

### Step 7: Locus of \( z \)
From the second case, we have two lines:
\[ y = \sqrt{3}x \quad \text{and} \quad y = -\sqrt{3}x \]
This represents a pair of straight lines through the origin.

### Conclusion
The locus of \( z \) is the pair of straight lines \( y = \sqrt{3}x \) and \( y = -\sqrt{3}x \), along with the origin \( (0,0) \).