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

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(B) Let $z = x + iy$. Then $\frac{2z+1}{iz+1} = \frac{2(x+iy)+1}{i(x+iy)+1} = \frac{(2x+1) + 2yi}{(1-y) + xi}$.Multiplying the numerator and denominat

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a straight line

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(B) Let $z = x + iy$. Then $\frac{2z+1}{iz+1} = \frac{2(x+iy)+1}{i(x+iy)+1} = \frac{(2x+1) + 2yi}{(1-y) + xi}$.Multiplying the numerator and denominator by the conjugate of the denominator $(1-y) - xi$:$\frac{((2x+1) + 2yi)((1-y) - xi)}{(1-y)^2 + x^2} = \frac{(2x+1)(1-y) + 2xy + i(2y(1-y) - x(2x+1))}{(1-y)^2 + x^2}$.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:$-x - 2y + 2 = 0$,which is $x + 2y - 2 = 0$.This represents 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 Argand plane is

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