The equation `|z-1|^(2)+|z+1|^(2)=2`, represent

Options

A. a circle of radius one unit
B. a straight line
C. the ordered pair (0,0)
D. none of these

Explanation

To solve the equation \( |z-1|^2 + |z+1|^2 = 2 \), we will 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: Rewrite the equation using the definition of modulus
The modulus of a complex number \( z = x + iy \) is given by \( |z| = \sqrt{x^2 + y^2} \). Therefore, we can express \( |z - 1| \) and \( |z + 1| \) as follows:

\[ |z - 1| = |(x - 1) + iy| = \sqrt{(x - 1)^2 + y^2} \]
\[ |z + 1| = |(x + 1) + iy| = \sqrt{(x + 1)^2 + y^2} \]

### Step 3: Substitute these into the original equation
Now we substitute these into the equation:

\[ |z - 1|^2 + |z + 1|^2 = 2 \]
This becomes:

\[ \sqrt{(x - 1)^2 + y^2}^2 + \sqrt{(x + 1)^2 + y^2}^2 = 2 \]

Which simplifies to:

\[ (x - 1)^2 + y^2 + (x + 1)^2 + y^2 = 2 \]

### Step 4: Expand and simplify the equation
Expanding the squares gives:

\[ (x^2 - 2x + 1 + y^2) + (x^2 + 2x + 1 + y^2) = 2 \]

Combining like terms results in:

\[ 2x^2 + 2y^2 + 2 = 2 \]

### Step 5: Further simplify the equation
Subtracting 2 from both sides gives:

\[ 2x^2 + 2y^2 = 0 \]

Dividing through by 2 yields:

\[ x^2 + y^2 = 0 \]

### Step 6: Analyze the equation
The equation \( x^2 + y^2 = 0 \) implies that both \( x \) and \( y \) must be zero, since the sum of squares is zero only when each square is zero.

Thus, we have:

\[ x = 0 \quad \text{and} \quad y = 0 \]

### Conclusion
The only solution is the point \( (0, 0) \), which corresponds to the complex number \( z = 0 \).

Therefore, the equation \( |z-1|^2 + |z+1|^2 = 2 \) represents the single point \( z = 0 \).

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