JEE MATHEMATICS MCQ: COMPLEX NUMBERS

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Analyzing +4 | -1

The locus of the complex number

z

in an argand plane satisfying the inequality

\log _{1 / 2}\left(\frac{|z-1|+4}{3|z-1|-2}\right) > 1\left(\right.

where,

\left.|z-1| \neq \frac{2}{3}\right)

View answer and explanation

Correct option: C

exterior of a circle

Option C (exterior of a circle) is correct because the inequality simplifies to

|z-1| > 3

, which represents the region outside a circle centered at 1 with radius 3.

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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}