The locus represented by |z-1|=|z+i| is:

Options

A. The circle of radius 1
B. An ellipse with foci at 1 and – i
C. A circle through the origin
D. Line passing through origin

Answer

To solve the problem of finding the locus represented by the equation |z - 1| = |z + i|, we can follow these steps:

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

### Step 2: Rewrite the equation
The equation becomes:
\[ |z - 1| = |z + i| \]
Substituting \( z \):
\[ | (x + iy) - 1 | = | (x + iy) + i | \]
This simplifies to:
\[ | (x - 1) + iy | = | x + (y + 1)i | \]

### Step 3: Calculate the moduli
Now, we calculate the moduli on both sides:
\[ \sqrt{(x - 1)^2 + y^2} = \sqrt{x^2 + (y + 1)^2} \]

### Step 4: Square both sides
To eliminate the square roots, we square both sides:
\[ (x - 1)^2 + y^2 = x^2 + (y + 1)^2 \]

### Step 5: Expand both sides
Expanding both sides gives:
\[ (x^2 - 2x + 1 + y^2) = (x^2 + y^2 + 2y + 1) \]

### Step 6: Simplify the equation
Now, we can simplify the equation:
- The \( x^2 \) terms cancel out.
- The \( y^2 \) terms also cancel out.
- The constant \( 1 \) on both sides cancels out.

This leaves us with:
\[ -2x = 2y \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ x + y = 0 \]

### Conclusion
The equation \( x + y = 0 \) represents a straight line that passes through the origin.

### Final Answer
The locus represented by \( |z - 1| = |z + i| \) is a line passing through the origin.

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