If ${\mathop{\rm Im}\nolimits} \left( {\frac{{z - 1}}{{2z + 1}}} \right) = - 4$, then locus of $z$ is

JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy

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Application +4 | -1

{\mathop\mathrm{Im}\nolimits} \left( {\frac{{z - 1}}{{2z + 1}}} \right) = - 4

, then locus of

z

View answer and explanation

Correct option: D

A Circle

Option D (A Circle) is correct because the imaginary part condition leads to an equation of the form

16{x^2} + 16{y^2} + 16x + 3y + 4 = 0

, which represents a circle.

Practice All Chapter MCQs →

{\left| {1 - i} \right|^x} = {2^x}

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z \neq \pm 1

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\left|\sum\limits_{r=1}^n z_r\right|

(z+a b)^3=a^3

a \neq 0

\alpha, \beta

\gamma

|\alpha-\beta|

{z^2} + {\left| z \right|^2} = 0,

z \in C

2x = - 1 + \sqrt 3 i

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