Consider the regions for complex number z A:1 _2 | z | - 1 _2 | z… - Vidyadip

MCQ

Consider the regions for complex number $z$

$A:\frac{1}{{\log_2 \left| z \right|}} - \frac{1}{{\log_2 \left| z \right| - 1}} - 1 < 0$ and $\left( {B:\operatorname{Im} \left( z \right) = 0} \right).$ The range of values of $Re(z)$
lying in the region $A \cap B$ is

A
$\left( { - \infty , - 1} \right) \cup \left( {1,\infty } \right)$
B
$\left( { - \infty , - 2} \right) \cup \left( { - 1,1} \right) \cup \left( {2,\infty } \right)$
✓
$\left( { - \infty , - 2} \right) \cup \left( { - 1,0} \right) \cup \left( {0,1} \right) \cup \left( {2,\infty } \right)$
D
$\left( { - \infty , - 2} \right) \cup \left( { - 1,0} \right) \cup \left( {2,\infty } \right)$

✓

Answer

Correct option: C.

$\left( { - \infty , - 2} \right) \cup \left( { - 1,0} \right) \cup \left( {0,1} \right) \cup \left( {2,\infty } \right)$

c
$z=x+i y$

$\mathrm{Imz}=0 \Rightarrow \mathrm{y}=0$

$\therefore \quad|z|=|x|$

$\frac{1}{\log _{2}|x|}-\frac{1}{\log _{2}|x|-1}<1$

$\Rightarrow \frac{-1}{\left(\log _{2}|\mathrm{x}|\right)\left(\log _{2}|\mathrm{x}|-1\right)}<1$

$\Rightarrow \frac{\left(\log _{2}|\mathrm{x}|\right)\left(\log _{2}|\mathrm{x}|-1\right)+1}{\left(\log _{2}|\mathrm{x}|\right)\left(\log _{2}|\mathrm{x}|-1\right)}>0$

Numerator $=t^{2}-t+1$ is $+v e\left(\text { where } \log _{2}|x|=t\right)$

$\therefore \quad\left(\log _{2}|x|\right)\left(\log _{2}|x|-1\right)>0$

${\log _{2}|x|<0} {\text { or } \log _{2}|x|>1} $

${0<|x|<1} {\text { or }|x|>2} $

$ - 1 < x < 1,x \ne 0\,\,\,\,\,\,\,\,\,\,\,or\,\,\,\,\,\,\,\,\,\,\,x < - 2\,\,or\,\,x > 2$

${x \in(-\infty,-2) \cup(-1,1) \cup(2, \infty),} {x \neq 0}$

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