Number of solutions of the equation `z^(2)+|z|^(2)=0`, where `z in C`, is

Options

A. 1
B. 2
C. 3
D. infinity many

Answer

To solve the equation \( z^2 + |z|^2 = 0 \) where \( z \in \mathbb{C} \), we will follow these steps:

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

### Step 2: Substitute \( z \) into the equation
We substitute \( z \) into the equation:
\[ z^2 + |z|^2 = 0 \]
This becomes:
\[ (x + iy)^2 + |x + iy|^2 = 0 \]

### Step 3: Calculate \( z^2 \) and \( |z|^2 \)
Calculating \( z^2 \):
\[ (x + iy)^2 = x^2 + 2xyi - y^2 = (x^2 - y^2) + 2xyi \]
Calculating \( |z|^2 \):
\[ |z|^2 = x^2 + y^2 \]

### Step 4: Combine the results
Now, substituting these results back into the equation:
\[ (x^2 - y^2) + 2xyi + (x^2 + y^2) = 0 \]
This simplifies to:
\[ (2x^2 - y^2) + 2xyi = 0 \]

### Step 5: Set real and imaginary parts to zero
For the equation to hold, both the real and imaginary parts must equal zero:
1. Real part: \( 2x^2 - y^2 = 0 \)
2. Imaginary part: \( 2xy = 0 \)

### Step 6: Solve the equations
From the imaginary part \( 2xy = 0 \), we have two cases:
- Case 1: \( x = 0 \)
- Case 2: \( y = 0 \)

#### Case 1: \( x = 0 \)
Substituting \( x = 0 \) into the real part:
\[ 2(0)^2 - y^2 = 0 \implies -y^2 = 0 \implies y = 0 \]
Thus, \( z = 0 \).

#### Case 2: \( y = 0 \)
Substituting \( y = 0 \) into the real part:
\[ 2x^2 - (0)^2 = 0 \implies 2x^2 = 0 \implies x = 0 \]
Thus, \( z = 0 \).

### Conclusion
In both cases, we find that the only solution is \( z = 0 \). Therefore, the number of solutions of the equation \( z^2 + |z|^2 = 0 \) is:

\[ \text{Number of solutions} = 1 \]

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