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

Options

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

Answer

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

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

### Step 2: Set up the equations
We need to express the conditions given in the problem:
1. \( |z - i| = |z + i| \)
2. \( |z - i| = |z + 1| \)

### Step 3: Calculate the moduli
1. For \( |z - i| \):
\[ |z - i| = |(x + iy) - i| = |x + i(y - 1)| = \sqrt{x^2 + (y - 1)^2} \]

2. For \( |z + i| \):
\[ |z + i| = |(x + iy) + i| = |x + i(y + 1)| = \sqrt{x^2 + (y + 1)^2} \]

3. For \( |z + 1| \):
\[ |z + 1| = |(x + iy) + 1| = |(x + 1) + iy| = \sqrt{(x + 1)^2 + y^2} \]

### Step 4: Set up the equations from the moduli
From the conditions, we have:
1. \( \sqrt{x^2 + (y - 1)^2} = \sqrt{x^2 + (y + 1)^2} \)
2. \( \sqrt{x^2 + (y - 1)^2} = \sqrt{(x + 1)^2 + y^2} \)

### Step 5: Square both sides to eliminate the square roots
1. Squaring the first equation:
\[ x^2 + (y - 1)^2 = x^2 + (y + 1)^2 \]
Simplifying gives:
\[ (y - 1)^2 = (y + 1)^2 \]
Expanding both sides:
\[ y^2 - 2y + 1 = y^2 + 2y + 1 \]
Cancelling \( y^2 + 1 \) from both sides:
\[ -2y = 2y \implies 4y = 0 \implies y = 0 \]

2. Squaring the second equation:
\[ x^2 + (y - 1)^2 = (x + 1)^2 + y^2 \]
Substituting \( y = 0 \):
\[ x^2 + (0 - 1)^2 = (x + 1)^2 + 0^2 \]
This simplifies to:
\[ x^2 + 1 = (x + 1)^2 \]
Expanding the right side:
\[ x^2 + 1 = x^2 + 2x + 1 \]
Cancelling \( x^2 + 1 \) from both sides:
\[ 0 = 2x \implies x = 0 \]

### Step 6: Conclusion
The only solution is \( x = 0 \) and \( y = 0 \), which corresponds to the complex number \( z = 0 + 0i = 0 \).

Thus, the number of complex numbers \( z \) that satisfy the given conditions is **1**.