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

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

### Step 2: Use the modulus condition
The modulus condition \( |z| = 2 \) gives us:
\[ |z| = \sqrt{x^2 + y^2} = 2 \]
Squaring both sides, we obtain:
\[ x^2 + y^2 = 4 \quad \text{(Equation 1)} \]

### Step 3: Find \( z^2 \)
Now, we calculate \( z^2 \):
\[ z^2 = (x + iy)^2 = x^2 + 2xyi - y^2 = (x^2 - y^2) + 2xyi \]
The real part of \( z^2 \) is:
\[ \text{Re}(z^2) = x^2 - y^2 \]

### Step 4: Set the real part to zero
According to the problem, we have:
\[ \text{Re}(z^2) = 0 \implies x^2 - y^2 = 0 \quad \text{(Equation 2)} \]
This implies:
\[ x^2 = y^2 \implies y = \pm x \]

### Step 5: Substitute \( y \) into Equation 1
Now, substituting \( y = x \) and \( y = -x \) into Equation 1:
1. For \( y = x \):
\[ x^2 + x^2 = 4 \implies 2x^2 = 4 \implies x^2 = 2 \implies x = \pm \sqrt{2} \]
Thus, \( y = \pm \sqrt{2} \). This gives us the points:
- \( (\sqrt{2}, \sqrt{2}) \)
- \( (\sqrt{2}, -\sqrt{2}) \)
- \( (-\sqrt{2}, \sqrt{2}) \)
- \( (-\sqrt{2}, -\sqrt{2}) \)

### Step 6: Count the solutions
From the above calculations, we find that there are 4 solutions:
1. \( z_1 = \sqrt{2} + i\sqrt{2} \)
2. \( z_2 = \sqrt{2} - i\sqrt{2} \)
3. \( z_3 = -\sqrt{2} + i\sqrt{2} \)
4. \( z_4 = -\sqrt{2} - i\sqrt{2} \)

### Final Answer
The number of solutions of the system of equations is **4**.

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