If P is the affix of z in the Argand diagram and P moves so that `(z - i)/( z - 1)` is always purely imaginary, then locus of z is

Options

A. circle, centre (2,2) radius 1/2
B. circle, centre (-1/2, -1/2) , radius `1 // sqrt(2)`
C. circle, centre (1/2,1/2) , radius `1//sqrt(2)`
D. none of these

Answer

To find the locus of the complex number \( z \) such that the expression \( \frac{z - i}{z - 1} \) is purely imaginary, we can follow these steps:

### Step 1: Let \( z = x + iy \)
We start by expressing \( z \) in terms of its real and imaginary parts:
\[ z = x + iy \]
where \( x \) is the real part and \( y \) is the imaginary part.

### Step 2: Substitute \( z \) into the expression
Substituting \( z \) into the expression \( \frac{z - i}{z - 1} \):
\[ \frac{(x + iy) - i}{(x + iy) - 1} = \frac{x + i(y - 1)}{(x - 1) + iy} \]

### Step 3: Multiply by the conjugate of the denominator
To simplify, we multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{(x + i(y - 1))((x - 1) - iy)}{((x - 1) + iy)((x - 1) - iy)} \]

### Step 4: Simplify the denominator
The denominator simplifies as follows:
\[ ((x - 1) + iy)((x - 1) - iy) = (x - 1)^2 + y^2 \]

### Step 5: Simplify the numerator
Now, simplifying the numerator:
\[ (x + i(y - 1))((x - 1) - iy) = x(x - 1) - xy + i(y - 1)(x - 1) + iy^2 \]
This gives:
\[ = x^2 - x - xy + i[(y - 1)(x - 1) + y^2] \]

### Step 6: Separate real and imaginary parts
Now we have:
\[ \frac{(x^2 - x - xy) + i[(y - 1)(x - 1) + y^2]}{(x - 1)^2 + y^2} \]
For this expression to be purely imaginary, the real part must be zero:
\[ x^2 - x - xy = 0 \]

### Step 7: Solve for \( y \)
Rearranging gives:
\[ x^2 - x = xy \implies y = \frac{x^2 - x}{x} \quad \text{(for } x \neq 0\text{)} \]
This simplifies to:
\[ y = x - 1 \]

### Step 8: Identify the locus
The equation \( y = x - 1 \) represents a straight line in the Argand diagram.

### Conclusion
Thus, the locus of \( z \) is a straight line given by the equation:
\[ y = x - 1 \]