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

JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy

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Application +4 | -1

\mathrm{Im}\left( {\frac{{2\;z + 1}}{{i\;z + 1}}} \right) = - 2,

then locus of

z

View answer and explanation

Correct option: A

A circle

Option A (A circle) is correct because the given condition on the imaginary part of the complex expression leads to an equation that represents a circle in the complex plane.

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