The number of jsolutions of the equation `z^(2)+barz=0,` is

Options

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

Answer

To find the number of solutions for the equation \( z^2 + \bar{z} = 0 \), we will follow these steps:

### Step 1: Substitute \( z \) with \( x + iy \)
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. The conjugate \( \bar{z} \) is \( x - iy \).

### Step 2: Rewrite the equation
Substituting \( z \) and \( \bar{z} \) into the equation gives:
\[ (x + iy)^2 + (x - iy) = 0 \]

### Step 3: Expand the equation
Expanding \( (x + iy)^2 \):
\[ x^2 + 2xyi - y^2 + x - iy = 0 \]
This simplifies to:
\[ (x^2 - y^2 + x) + (2xy - y)i = 0 \]

### Step 4: Set real and imaginary parts to zero
For the equation to hold, both the real and imaginary parts must equal zero:
1. Real part: \( x^2 - y^2 + x = 0 \) (Equation 1)
2. Imaginary part: \( 2xy - y = 0 \) (Equation 2)

### Step 5: Solve the imaginary part
From Equation 2, factor out \( y \):
\[ y(2x - 1) = 0 \]
This gives us two cases:
1. \( y = 0 \)
2. \( 2x - 1 = 0 \) which implies \( x = \frac{1}{2} \)

### Step 6: Solve for \( x \) when \( y = 0 \)
Substituting \( y = 0 \) into Equation 1:
\[ x^2 + x = 0 \]
Factoring gives:
\[ x(x + 1) = 0 \]
Thus, \( x = 0 \) or \( x = -1 \).

### Step 7: Solve for \( y \) when \( x = \frac{1}{2} \)
Substituting \( x = \frac{1}{2} \) into Equation 1:
\[ \left(\frac{1}{2}\right)^2 - y^2 + \frac{1}{2} = 0 \]
This simplifies to:
\[ \frac{1}{4} - y^2 + \frac{1}{2} = 0 \implies -y^2 + \frac{3}{4} = 0 \implies y^2 = \frac{3}{4} \]
Thus, \( y = \pm \frac{\sqrt{3}}{2} \).

### Step 8: Compile the solutions
The solutions we have are:
1. \( (0, 0) \)
2. \( (-1, 0) \)
3. \( \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \)
4. \( \left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) \)

### Conclusion
We have a total of 4 solutions:
1. \( z = 0 \)
2. \( z = -1 \)
3. \( z = \frac{1}{2} + i\frac{\sqrt{3}}{2} \)
4. \( z = \frac{1}{2} - i\frac{\sqrt{3}}{2} \)

Thus, the number of solutions to the equation \( z^2 + \bar{z} = 0 \) is **4**.