If $I m (\frac{2 z + 1}{i z + 1}) = - 4,$ then the locus of the point representing $z$ in the complex plane is

a straight line

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a parabola

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a circle

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a hyperbola

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Solution

The correct option is C a circle
Let
$z = x + i y$
Hence
$\frac{2 x + 1 + i 2 y}{(1 - y) + i x} = \frac{(2 x + 1 + i 2 y) ((1 - y) - i x))}{(1 - y)^{2} + x^{2}}$
We get the imaginary part as
$\frac{- x (2 x + 1) + 2 y (1 - y)}{(1 - y)^{2} + x^{2}} = - 4$
Hence
$- 2 y^{2} - 2 x^{2} - x + 2 y = - 4 (x^{2} + (y - 1)^{2})$
$2 x^{2} + 2 y^{2} - 2 y + x = 4 x^{2} + 4 y^{2} - 8 y + 4$
$2 x^{2} + 2 y^{2} - 6 y - x + 4 = 0$
Hence an equation of a circle.

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Similar questions

Im (\frac{2 z + 1}{i z + 1}) = - 3

z

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:

Im (\frac{2 z + 1}{i z + 1}) = - 3

z

\frac{2 z + 1}{i z + 1}

- 2

z

z = x + i y

\frac{2 z + 1}{i z + 1}

- 2

z

x + 2 y - 2 = 0

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