The number of solutions of the system of equations `"Re(z^(2))=0, |z|=2`, is

Options

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

Answer

To find the number of solutions for the system of equations given by \( \text{Re}(z^2) = 0 \) and \( |z| = 2 \), we can follow these steps:

### Step 1: Define the complex number
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers.

### Step 2: Compute \( z^2 \)
Calculating \( z^2 \):
\[ z^2 = (x + iy)^2 = x^2 + 2xyi - y^2 = (x^2 - y^2) + 2xyi \]
Here, the real part of \( z^2 \) is \( x^2 - y^2 \) and the imaginary part is \( 2xy \).

### Step 3: Set the real part to zero
According to the first equation \( \text{Re}(z^2) = 0 \):
\[ x^2 - y^2 = 0 \]
This implies:
\[ x^2 = y^2 \quad \Rightarrow \quad y = \pm x \]

### Step 4: Use the modulus condition
According to the second equation \( |z| = 2 \):
\[ |z| = \sqrt{x^2 + y^2} = 2 \]
Squaring both sides gives:
\[ x^2 + y^2 = 4 \]

### Step 5: Substitute \( y \) in terms of \( x \)
Substituting \( y = x \) into the modulus equation:
\[ x^2 + x^2 = 4 \quad \Rightarrow \quad 2x^2 = 4 \quad \Rightarrow \quad x^2 = 2 \quad \Rightarrow \quad x = \pm \sqrt{2} \]
Thus, when \( y = x \):
- If \( x = \sqrt{2} \), then \( y = \sqrt{2} \).
- If \( x = -\sqrt{2} \), then \( y = -\sqrt{2} \).

Now substituting \( y = -x \):
\[ x^2 + (-x)^2 = 4 \quad \Rightarrow \quad 2x^2 = 4 \quad \Rightarrow \quad x^2 = 2 \quad \Rightarrow \quad x = \pm \sqrt{2} \]
Thus, when \( y = -x \):
- If \( x = \sqrt{2} \), then \( y = -\sqrt{2} \).
- If \( x = -\sqrt{2} \), then \( y = \sqrt{2} \).

### Step 6: List all solutions
The solutions for \( z \) are:
1. \( z = \sqrt{2} + \sqrt{2}i \)
2. \( z = \sqrt{2} - \sqrt{2}i \)
3. \( z = -\sqrt{2} + \sqrt{2}i \)
4. \( z = -\sqrt{2} - \sqrt{2}i \)

### Conclusion
Thus, the number of solutions to the system of equations is **4**.

---