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

Options

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

Answer

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

### Step 1: Set up the equations
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. The given conditions can be expressed as:
1. \( |z - 1| = |z + 1| \)
2. \( |z - 1| = |z - i| \)

### Step 2: Analyze the first equation
From the first condition, we have:
\[ |z - 1| = |z + 1| \]
This translates to:
\[ |x - 1 + iy| = |x + 1 + iy| \]
Calculating the moduli gives:
\[ \sqrt{(x - 1)^2 + y^2} = \sqrt{(x + 1)^2 + y^2} \]
Squaring both sides, we get:
\[ (x - 1)^2 + y^2 = (x + 1)^2 + y^2 \]
The \( y^2 \) terms cancel out, leading to:
\[ (x - 1)^2 = (x + 1)^2 \]
Expanding both sides:
\[ x^2 - 2x + 1 = x^2 + 2x + 1 \]
Simplifying gives:
\[ -2x = 2x \implies 4x = 0 \implies x = 0 \]

### Step 3: Analyze the second equation
Now, substituting \( x = 0 \) into the second condition:
\[ |z - 1| = |z - i| \]
This translates to:
\[ |0 - 1 + iy| = |0 - i + iy| \]
Calculating the moduli gives:
\[ | -1 + iy | = | -i + iy | \]
This can be expressed as:
\[ \sqrt{(-1)^2 + y^2} = \sqrt{(-y)^2 + y^2} \]
Simplifying both sides leads to:
\[ \sqrt{1 + y^2} = \sqrt{y^2 + 1} \]
Squaring both sides:
\[ 1 + y^2 = y^2 + 1 \]
This is always true, meaning \( y \) can take any value.

### Step 4: Conclusion
Since \( x = 0 \) and \( y \) can be any real number, the complex numbers \( z \) can be expressed as:
\[ z = 0 + iy \]
This means all purely imaginary numbers are solutions.

### Final Answer
The number of complex numbers \( z \) such that \( |z - 1| = |z + 1| = |z - i| \) is infinite, as \( y \) can take any real value.

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