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If the imaginary part of $\frac{2 z+1}{i z+1}$ is -2 , then the locus of the point representing $\mathrm{z}$ in the Argand plane is
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Verified Answer
The correct answer is:
a straight line
$$
\begin{aligned}
& \frac{2 z+1}{i z+1}=\frac{(2 x+1)+2 y i}{(1-y)+x i} \\
= & \frac{(2 x+1)+2 y i}{(1-y)+x i} \times \frac{(1-y)-x i}{(1-y)-x i} \\
\because & \text { Imaginary part }=-2 \\
\therefore \quad & \frac{-x(2 x+1)+2 y(1-y)}{(1-y)^2+x^2}=-2 \\
\Rightarrow & -2 x^2-x+2 y-2 y^2=-2-2 y^2+4 y-2 x^2 \\
\Rightarrow & -x-2 y+2=0 \\
\Rightarrow & x+2 y-2=0
\end{aligned}
$$
which is a straight line.
\begin{aligned}
& \frac{2 z+1}{i z+1}=\frac{(2 x+1)+2 y i}{(1-y)+x i} \\
= & \frac{(2 x+1)+2 y i}{(1-y)+x i} \times \frac{(1-y)-x i}{(1-y)-x i} \\
\because & \text { Imaginary part }=-2 \\
\therefore \quad & \frac{-x(2 x+1)+2 y(1-y)}{(1-y)^2+x^2}=-2 \\
\Rightarrow & -2 x^2-x+2 y-2 y^2=-2-2 y^2+4 y-2 x^2 \\
\Rightarrow & -x-2 y+2=0 \\
\Rightarrow & x+2 y-2=0
\end{aligned}
$$
which is a straight line.
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