Question

If $z = x + iy , x , y \in R$ and the imaginary part of $\frac{\bar{z} - 1}{\bar{z} - i}$ is $1$ then the locus of $z$ is

• A. $x + y + 1 = 0$
• B. $x + y + 1 = 0, (x, y) \neq (0, - 1)$
• C. $x^2 + y^2 - x +3y + 2 = 0$
• D. $x^2 + y^2 - x + 3y + 2 = 0 , (x , y) \neq (0 , - 1)$

Answer 1

Answer: (D)

$x^2 + y^2 - x + 3y + 2 = 0 , (x , y) \neq (0 , - 1)$

Answer 2

The correct answer is D:x^2+y^2-x+3y+2=0,(x,y)$$≠(0,-1)
If $z=x +i y$, then
\frac{\bar{z}-1}{\bar{z}-i}=\frac{x-i y-1}{x-i y-i} \times \frac{x +i(y+ 1 )}{x +i(y+ 1 )}$$=\frac{[x(x-1)+y(y+1)]+i[(y+1)(x-1)-x y]}{x^{2}+(y+1)^{2}}$$\therefore Im \left(\frac{\bar{z}-1}{\bar{z}-i}\right)=\frac{x y-y+x-1-x y}{x^{2}+(y+1)^{2}}=1 (given)
$\Rightarrow x^{2}+y^{2}-x+3 y+2=0,(x, y) \neq(0,-1)$