Question

If the point \(P\) represents a complex number \(z\) in the Argand diagram and \[ \frac{z-i}{z-1} \] is always purely imaginary, then the locus of \(P\) is

• A. Circle with centre \( \left(\frac12,\frac12\right) \) and radius \( \frac{1}{\sqrt2} \)
• B. Circle with centre \( \left(-\frac12,-\frac12\right) \) and radius \( \frac{1}{\sqrt2} \)
• C. Circle with centre \( \left(\frac12,\frac12\right) \) and radius \( \frac{1}{\sqrt2} \) except the points \((0,1)\) and \((1,0)\)
• D. Circle with centre \( \left(-\frac12,-\frac12\right) \) and radius \( \frac12 \) except the point \((1,0)\) \bigskip

Hint:

Whenever a complex expression is stated to be purely imaginary, set its real part equal to zero. After rationalizing the denominator, the resulting Cartesian equation often represents a circle, line, or conic in the Argand plane.

Answer 1

The Correct Option is A

Concept: A complex number is purely imaginary when its real part is zero. Convert the given expression into Cartesian form and equate its real part to zero.

Step 1: Let \(z=x+iy\). Then \(\dfrac{z-i}{z-1}=\dfrac{(x+i(y-1))((x-1)-iy)}{(x-1)^2+y^2}\).

Step 2: The real part of the numerator is \(x(x-1)+y(y-1)\). Hence \(x(x-1)+y(y-1)=0\).

Step 3: Completing the squares gives \(\left(x-\frac12\right)^2+\left(y-\frac12\right)^2=\frac12\). Thus the required locus is \(\boxed{\left(x-\frac12\right)^2+\left(y-\frac12\right)^2=\frac12}\).