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If the real part of $\frac{\bar{z}+2}{\bar{z}-1}$ is 4 , then show that the locus of the point representing $z$ in the complex plane is a circle.
Solution:
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Verified Answer
Let $\quad z=x+i y$
Now, $\frac{\bar{z}+2}{\bar{z}-1}=\frac{x-i y+2}{x-i y-1}$
$=\frac{[(x+2)-i y][(x-1)+i y]}{[(x-1)-i y][(x-1)+i y]} \quad$ [Rationalizing $]$
$$
=\frac{\begin{array}{l}
(x-1)(x+2)-i y(x-1) \\
+i y(x+2)+y^2
\end{array}}{(x-1)^2+y^2}
$$
$$
=\frac{\begin{array}{l}
(x-1)(x+2)+y^2 \\
+i[(x+2) y-(x-1) y]
\end{array}}{(x-1)^2+y^2} \quad\left[\because-i^2=1\right]
$$
Taking real part, $\frac{(x-1)(x+2)+y^2}{(x-1)^2+y^2}=4$
$$
\Rightarrow x^2-x+2 x-2+y^2=4\left(x^2-2 x+1+y^2\right)
$$
$\Rightarrow \quad 3 x^2+3 y^2-9 x+6=0$, which represents a circle. Hence, $z$ lies on the circle.
Now, $\frac{\bar{z}+2}{\bar{z}-1}=\frac{x-i y+2}{x-i y-1}$
$=\frac{[(x+2)-i y][(x-1)+i y]}{[(x-1)-i y][(x-1)+i y]} \quad$ [Rationalizing $]$
$$
=\frac{\begin{array}{l}
(x-1)(x+2)-i y(x-1) \\
+i y(x+2)+y^2
\end{array}}{(x-1)^2+y^2}
$$
$$
=\frac{\begin{array}{l}
(x-1)(x+2)+y^2 \\
+i[(x+2) y-(x-1) y]
\end{array}}{(x-1)^2+y^2} \quad\left[\because-i^2=1\right]
$$
Taking real part, $\frac{(x-1)(x+2)+y^2}{(x-1)^2+y^2}=4$
$$
\Rightarrow x^2-x+2 x-2+y^2=4\left(x^2-2 x+1+y^2\right)
$$
$\Rightarrow \quad 3 x^2+3 y^2-9 x+6=0$, which represents a circle. Hence, $z$ lies on the circle.
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