If $\left| {\frac{{z - 2}}{{z + 2}}} \right| = \frac{{\rm{\pi }}}{6}$, then the locus of z is

JEE MATHEMATICS MCQ: COMPLEX NUMBERS

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Application +4 | -1

\left| {\frac{{z - 2}}{{z + 2}}} \right| = \frac{\mathrm{\pi }}{6}

, then the locus of z is

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Correct option: B

A circle

Option B (A circle) is correct because the given equation represents a circle in the complex plane.

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

|z|=1

z \neq \pm 1

z, \frac{z}{1-z^2}

\left|z_r-r\right| \leq r, \forall r=1,2,3, \ldots, n

\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

{\left( {1 - {x^2} + x} \right)^6} - {\left( {1 - x + {x^2}} \right)^6}