If \(z=x+i y, x, y \in R\) and the imaginary part of \(\frac{\bar{z}-1}{\bar{z}-i}\) is 1 , then the locus…
If \(z=x+i y, x, y \in R\) and the imaginary part of \(\frac{\bar{z}-1}{\bar{z}-i}\) is 1 , then the locus of \(z\) is
- \(x+y+1=0\)
- \(x+y+1=0,(x, y) \neq(0,-1)\)
- \(x^2+y^2-x+3 y+2=0\)
- \(x^2+y^2-x+3 y+2=0,(x, y) \neq(0,-1)\)
Solution
If \(z=x+i y\), then
\(\begin{aligned}
& \frac{\bar{z}-1}{\bar{z}-i}=\frac{x-i y-1}{x-i y-i} \times \frac{x+i(y+1)}{x+i(y+1)} \\
= & \frac{[x(x-1)+y(y+1)]+i[(y+1)(x-1)-x y]}{x^2+(y+1)^2} \\
\therefore & \quad \operatorname{Im}\left(\frac{\bar{z}-1}{\bar{z}-i}\right)=\frac{x y-y+x-1-x y}{x^2+(y+1)^2}=1 \quad \text{(given)} \\
\Rightarrow & x^2+y^2-x+3 y+2=0,(x, y) \neq(0,-1)
\end{aligned}\)
Hence, option (4) is correct.
Asked in: AP EAMCET 2019 (20 Apr Shift 1)
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