The number of roots of the equation `z^(2) = 2 bar(z)` is `:`

Options

A. A. 2
B. B. 3
C. C. 4
D. D. 0

Answer

To solve the equation \( z^2 = 2 \bar{z} \), where \( z \) is a complex number, we will follow these steps:

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

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

### Step 3: Expand both sides
Expanding the left-hand side:
\[ x^2 + 2xyi - y^2 = 2x - 2yi \]

### Step 4: Separate real and imaginary parts
Now, equate the real and imaginary parts:
1. Real part: \( x^2 - y^2 = 2x \)
2. Imaginary part: \( 2xy = -2y \)

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

### Step 6: Case 1: \( y = 0 \)
Substituting \( y = 0 \) into the real part equation:
\[ x^2 - 0 = 2x \implies x^2 - 2x = 0 \]
Factoring gives:
\[ x(x - 2) = 0 \]
Thus, \( x = 0 \) or \( x = 2 \).

So, we have two roots:
1. \( z = 0 + 0i = 0 \)
2. \( z = 2 + 0i = 2 \)

### Step 7: Case 2: \( x = -1 \)
Substituting \( x = -1 \) into the real part equation:
\[ (-1)^2 - y^2 = 2(-1) \implies 1 - y^2 = -2 \implies y^2 = 3 \]
Thus, \( y = \sqrt{3} \) or \( y = -\sqrt{3} \).

So, we have two more roots:
1. \( z = -1 + \sqrt{3}i \)
2. \( z = -1 - \sqrt{3}i \)

### Step 8: Conclusion
In total, we have the following roots:
1. \( z = 0 \)
2. \( z = 2 \)
3. \( z = -1 + \sqrt{3}i \)
4. \( z = -1 - \sqrt{3}i \)

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

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