If the imaginary part of frac{2z + 1}{iz + 1} | vedclass

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(B) Let $z = x + iy$. Then $\frac{2z + 1}{iz + 1} = \frac{2(x + iy) + 1}{i(x + iy) + 1} = \frac{(2x + 1) + 2iy}{(1 - y) + ix}$.Multiplying the numerat

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$A$ straight line

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(B) Let $z = x + iy$. Then $\frac{2z + 1}{iz + 1} = \frac{2(x + iy) + 1}{i(x + iy) + 1} = \frac{(2x + 1) + 2iy}{(1 - y) + ix}$.Multiplying the numerator and denominator by the conjugate of the denominator $(1 - y) - ix$:$= \frac{[(2x + 1) + 2iy][(1 - y) - ix]}{(1 - y)^2 + x^2} = \frac{(2x + 1)(1 - y) + 2xy + i[2y(1 - y) - x(2x + 1)]}{(1 - y)^2 + x^2}$.The imaginary part is given as $-2$:$\frac{2y - 2y^2 - 2x^2 - x}{(1 - y)^2 + x^2} = -2$.$2y - 2y^2 - 2x^2 - x = -2((1 - y)^2 + x^2) = -2(1 - 2y + y^2 + x^2) = -2 + 4y - 2y^2 - 2x^2$.Simplifying the equation:$2y - 2y^2 - 2x^2 - x = -2 + 4y - 2y^2 - 2x^2$.$-x - 2y = -2$,which simplifies to $x + 2y - 2 = 0$.This is the equation of a straight line.

If the imaginary part of $\frac{2z + 1}{iz + 1}$ is $-2$,then the locus of the point representing $z$ in the complex plane is

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