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

Options

A. 1
B. 2
C. `infty`
D. 0

Answer

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

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

### Step 2: Write the equations based on the modulus
We can express the given conditions in terms of \( x \) and \( y \):
1. \( |z - 1| = |z + 1| \)
2. \( |z - 1| = |z - i| \)

### Step 3: Expand the first equation
The first condition \( |z - 1| = |z + 1| \) translates to:
\[ |x + iy - 1| = |x + iy + 1| \]
This simplifies to:
\[ |(x - 1) + iy| = |(x + 1) + iy| \]
Calculating the modulus gives:
\[ \sqrt{(x - 1)^2 + y^2} = \sqrt{(x + 1)^2 + y^2} \]
Squaring both sides:
\[ (x - 1)^2 + y^2 = (x + 1)^2 + y^2 \]
The \( y^2 \) terms cancel out, leading to:
\[ (x - 1)^2 = (x + 1)^2 \]

### Step 4: Solve the equation
Expanding both sides:
\[ x^2 - 2x + 1 = x^2 + 2x + 1 \]
Subtracting \( x^2 + 1 \) from both sides:
\[ -2x = 2x \]
This simplifies to:
\[ -4x = 0 \implies x = 0 \]

### Step 5: Substitute \( x \) into the second equation
Now substituting \( x = 0 \) into the second condition \( |z - 1| = |z - i| \):
\[ |0 + iy - 1| = |0 + iy - i| \]
This translates to:
\[ |(-1) + iy| = |iy - i| \]
Calculating the moduli:
\[ \sqrt{(-1)^2 + y^2} = \sqrt{y^2 + 1} \]
This simplifies to:
\[ \sqrt{1 + y^2} = \sqrt{y^2 + 1} \]
Since both sides are equal, this equation holds for all \( y \).

### Step 6: Conclusion
Since \( x = 0 \) and \( y \) can take any real value, the complex numbers satisfying the conditions are of the form \( z = iy \) where \( y \) is any real number. Therefore, there are infinitely many complex numbers \( z \) that satisfy the given conditions.

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