If text{Im} left( frac{2z + 1}{iz + 1} right) | vedclass

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(A) 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 numerato

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a circle

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(A) 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 $\frac{2y - 2y^2 - 2x^2 - x}{(1 - y)^2 + x^2} = -3$.$2y - 2y^2 - 2x^2 - x = -3((1 - y)^2 + x^2)$.$2y - 2y^2 - 2x^2 - x = -3(1 - 2y + y^2 + x^2)$.$2y - 2y^2 - 2x^2 - x = -3 + 6y - 3y^2 - 3x^2$.Rearranging terms: $x^2 + y^2 - x - 4y + 3 = 0$.This is the equation of a circle.

If $\text{Im} \left( \frac{2z + 1}{iz + 1} \right) = -3$,then the locus of $z$ is :-

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