The number of complex numbers z such that `|z-1|=|z+1|=|z-i|` is

Options

A. 0
B. 1
C. 2
D. 3

Answer

To find the number of complex numbers \( z \) such that \( |z - 1| = |z + 1| = |z - i| \), we can follow these steps:

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

### Step 2: Write the equations for the moduli
We have three equations based on the given conditions:
1. \( |z - 1| = |z + 1| \)
2. \( |z - 1| = |z - i| \)
3. \( |z + 1| = |z - i| \)

### Step 3: Solve the first equation
From the first equation:
\[ |z - 1| = |z + 1| \]
This translates to:
\[ \sqrt{(x - 1)^2 + y^2} = \sqrt{(x + 1)^2 + y^2} \]
Squaring both sides gives:
\[ (x - 1)^2 + y^2 = (x + 1)^2 + y^2 \]
Cancelling \( y^2 \) from both sides:
\[ (x - 1)^2 = (x + 1)^2 \]
Expanding both sides:
\[ x^2 - 2x + 1 = x^2 + 2x + 1 \]
Cancelling \( x^2 + 1 \) from both sides:
\[ -2x = 2x \]
This simplifies to:
\[ 4x = 0 \implies x = 0 \]

### Step 4: Substitute \( x = 0 \) into the second equation
Now, substituting \( x = 0 \) into the second equation:
\[ |z - 1| = |z - i| \]
This translates to:
\[ |0 + iy - 1| = |0 + iy - i| \]
Which simplifies to:
\[ \sqrt{(-1)^2 + y^2} = \sqrt{y^2 + (-1)^2} \]
Both sides are equal, so this condition holds true for any \( y \).

### Step 5: Solve the third equation
Now, substituting \( x = 0 \) into the third equation:
\[ |z + 1| = |z - i| \]
This translates to:
\[ |0 + iy + 1| = |0 + iy - i| \]
Which simplifies to:
\[ \sqrt{1^2 + y^2} = \sqrt{y^2 + (-1)^2} \]
Again, both sides are equal, so this condition also holds true for any \( y \).

### Step 6: Conclusion
Since \( x = 0 \) and \( y \) can take any real value, the set of complex numbers \( z \) that satisfy the conditions is of the form:
\[ z = 0 + iy \quad \text{for any real } y \]
This represents a vertical line in the complex plane along the imaginary axis.

Thus, the number of complex numbers \( z \) satisfying the conditions is infinite.

### Final Answer
The number of complex numbers \( z \) such that \( |z - 1| = |z + 1| = |z - i| \) is **infinite**.