The locus of a point z represented by the equation `|z-1|=|z-i|` on the argand plane is `("where, "z in C, I = sqrt(-1))`

Options

A. a circle of radius 1
B. an ellipse with foci at 1 and `-i`
C. a line passing through the origin
D. a circle on the line joining 1 and `-i` as diameter

Answer

To find the locus of the point \( z \) represented by the equation \( |z - 1| = |z - i| \) on the Argand plane, we can follow these steps:

### Step 1: Express \( z \) in terms of its real and imaginary parts
Let \( z = x + iy \), where \( x \) is the real part and \( y \) is the imaginary part.

### Step 2: Rewrite the equation using the definition of modulus
The equation \( |z - 1| = |z - i| \) can be rewritten as:
\[ | (x + iy) - 1 | = | (x + iy) - i | \]
This simplifies to:
\[ | (x - 1) + iy | = | x + (y - 1)i | \]

### Step 3: Calculate the moduli
The modulus of a complex number \( a + bi \) is given by \( \sqrt{a^2 + b^2} \). Therefore, we have:
\[ \sqrt{(x - 1)^2 + y^2} = \sqrt{x^2 + (y - 1)^2} \]

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

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

### Step 6: Simplify the equation
Now, we can simplify the equation:
\[ x^2 - 2x + 1 + y^2 = x^2 + y^2 - 2y + 1 \]
Cancelling \( x^2 \), \( y^2 \), and \( 1 \) from both sides results in:
\[ -2x = -2y \]
This simplifies to:
\[ x = y \]

### Conclusion
The locus of the point \( z \) is the line \( y = x \) in the Argand plane.