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

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(C) Let $z = x + iy$. Substituting $z$ into the expression: $\frac{2z+1}{iz+1} = \frac{2(x+iy)+1}{i(x+iy)+1} = \frac{(2x+1) + i(2y)}{(1-y) + ix}$. To

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a straight line

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(C) Let $z = x + iy$. Substituting $z$ into the expression: $\frac{2z+1}{iz+1} = \frac{2(x+iy)+1}{i(x+iy)+1} = \frac{(2x+1) + i(2y)}{(1-y) + ix}$. To rationalize,multiply the numerator and denominator by the conjugate of the denominator,$(1-y) - ix$: $\frac{[(2x+1) + i(2y)][(1-y) - ix]}{(1-y)^2 + x^2} = \frac{(2x+1)(1-y) + 2xy + i[2y(1-y) - x(2x+1)]}{(1-y)^2 + x^2}$. The imaginary part is given as $-2$: $\frac{2y - 2y^2 - 2x^2 - x}{(1-y)^2 + x^2} = -2$. $2y - 2y^2 - 2x^2 - x = -2(1 - 2y + y^2 + x^2)$. $2y - 2y^2 - 2x^2 - x = -2 + 4y - 2y^2 - 2x^2$. Simplifying the equation: $-x - 2y = -2$,or $x + 2y - 2 = 0$. This is the equation of a straight line.

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

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