The number of solutions of `z^(2) + |z| = 0` is

Options

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

Answer

To solve the equation \( z^2 + |z| = 0 \) for the number of solutions, we will follow these steps:

### Step 1: Rewrite \( z \) in terms of its real and imaginary parts
Let \( z = x + iy \), where \( x \) is the real part and \( y \) is the imaginary part of the complex number \( z \).

### Step 2: Substitute \( z \) into the equation
Substituting \( z \) into the equation gives:
\[ (x + iy)^2 + |x + iy| = 0 \]
Calculating \( (x + iy)^2 \):
\[ (x + iy)^2 = x^2 - y^2 + 2xyi \]
Calculating the modulus \( |z| \):
\[ |z| = \sqrt{x^2 + y^2} \]
Thus, the equation becomes:
\[ x^2 - y^2 + 2xyi + \sqrt{x^2 + y^2} = 0 \]

### Step 3: Separate real and imaginary parts
For the equation to hold, both the real part and the imaginary part must equal zero:
1. Real part: \( x^2 - y^2 + \sqrt{x^2 + y^2} = 0 \)
2. Imaginary part: \( 2xy = 0 \)

### Step 4: Solve the imaginary part
From \( 2xy = 0 \), we have two cases:
1. \( x = 0 \)
2. \( y = 0 \)

### Case 1: \( x = 0 \)
Substituting \( x = 0 \) into the real part:
\[ 0 - y^2 + \sqrt{0 + y^2} = 0 \implies -y^2 + |y| = 0 \]
This gives us two subcases:
- If \( y \geq 0 \), then \( -y^2 + y = 0 \) leads to \( y(y - 1) = 0 \) which gives \( y = 0 \) or \( y = 1 \).
- If \( y < 0 \), then \( -y^2 - y = 0 \) leads to \( y(y + 1) = 0 \) which gives \( y = 0 \) or \( y = -1 \).

Thus, from \( x = 0 \), we have the solutions:
- \( (0, 0) \)
- \( (0, 1) \)
- \( (0, -1) \)

### Case 2: \( y = 0 \)
Substituting \( y = 0 \) into the real part:
\[ x^2 - 0 + \sqrt{x^2 + 0} = 0 \implies x^2 + |x| = 0 \]
This gives us:
- If \( x \geq 0 \), then \( x^2 + x = 0 \) leads to \( x(x + 1) = 0 \) which gives \( x = 0 \) or \( x = -1 \).
- If \( x < 0 \), then \( x^2 - x = 0 \) leads to \( x(x - 1) = 0 \) which gives \( x = 0 \) or \( x = 1 \).

Thus, from \( y = 0 \), we have the solutions:
- \( (0, 0) \)
- \( (1, 0) \)
- \( (-1, 0) \)

### Step 5: Compile all solutions
From both cases, we have the following solutions:
1. \( (0, 0) \)
2. \( (0, 1) \)
3. \( (0, -1) \)
4. \( (1, 0) \)
5. \( (-1, 0) \)

### Conclusion
The total number of distinct solutions is 5.