Given that `abs(z+i)=abs(z-i)=abs(z-1)` and `z=x+iy` find number of pairs (x,y)

Options

A. One
B. Two
C. Three
D. Four

Answer

To solve the problem given that \( |z + i| = |z - i| = |z - 1| \) where \( z = x + iy \), we will follow these steps:

### Step 1: Set up the equations
We start with the condition given in the problem:
1. \( |z + i| = |z - i| \)
2. \( |z + i| = |z - 1| \)

### Step 2: Express \( z \) in terms of \( x \) and \( y \)
Substituting \( z = x + iy \) into the equations:
1. \( |(x + iy) + i| = |(x + iy) - i| \)
- This simplifies to \( |x + i(y + 1)| = |x + i(y - 1)| \)

2. \( |(x + iy) + i| = |(x + iy) - 1| \)
- This simplifies to \( |x + i(y + 1)| = |(x - 1) + iy| \)

### Step 3: Solve the first equation
From the first equation \( |x + i(y + 1)| = |x + i(y - 1)| \):
- The magnitudes give us:
\[ \sqrt{x^2 + (y + 1)^2} = \sqrt{x^2 + (y - 1)^2} \]
- Squaring both sides:
\[ x^2 + (y + 1)^2 = x^2 + (y - 1)^2 \]
- Simplifying this, we get:
\[ (y + 1)^2 = (y - 1)^2 \]
- Expanding both sides:
\[ y^2 + 2y + 1 = y^2 - 2y + 1 \]
- Canceling \( y^2 + 1 \) from both sides:
\[ 2y = -2y \]
- Thus, \( 4y = 0 \) or \( y = 0 \).

### Step 4: Solve the second equation
Now substituting \( y = 0 \) into the second equation:
\[ |(x + i(0 + 1))| = |(x - 1) + i(0)| \]
This simplifies to:
\[ |x + i| = |x - 1| \]
- The magnitudes give us:
\[ \sqrt{x^2 + 1^2} = |x - 1| \]
- Squaring both sides:
\[ x^2 + 1 = (x - 1)^2 \]
- Expanding the right side:
\[ x^2 + 1 = x^2 - 2x + 1 \]
- Canceling \( x^2 + 1 \) from both sides:
\[ 0 = -2x \]
- Thus, \( x = 0 \).

### Conclusion
The only solution we have is \( (x, y) = (0, 0) \). Therefore, the number of pairs \( (x, y) \) that satisfy the given conditions is **1**.