Question

If ${{\log }_{\tan {{30}^{\circ }}}}\left( \dfrac{2{{\left| z \right|}^{2}}+2\left| z \right|-3}{\left| z \right|+1} \right)<-2$ then $$$$
A. \left| z \right|<\dfrac{3}{2}$$$$$ B. \left| z \right|>\dfrac{3}{2} C. $\left| z \right|>2
D. $\left| z \right|<2$$$$$

Answer 1

begin with the logarithmic identity of in equality to make the given inequality free of logarithm. Then proceed using the property of modulus to get to a quadratic inequality. Use the rule of product of two numbers to get the required value.$$$$
We know that any complex number $z=x+iy$ where $x$and $y$ are real numbers then the modulus of a complex x number is defined as $\left| z \right|=\left| x\pm iy \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$ which signifies the distance of the complex number from the origin in the complex plane and is always positive. The conjugate of $z$ is given by $\overline{z}=x-iy$. We can deduce that$z\overline{z}=\left( x+iy \right)\left( x-iy \right)={{x}^{2}}+{{y}^{2}}$ which is a real positive number.

Complete step-by-step answer:
The given expression is ${{\log }_{\tan {{30}^{\circ }}}}\left( \dfrac{2{{\left| z \right|}^{2}}+2\left| z \right|-3}{\left| z \right|+1} \right) < -2$. We know from the trigonometry that $\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}$. We also know from logarithm that the inequality ${{\log }_{a}}n < b \Rightarrow n > {{b}^{a}}$ will change sign when the base of logarithm $a$ lies between 0 and 1 . We use these to proceed,

& {{\log }_{\dfrac{1}{\sqrt{3}}}}\left( \dfrac{2{{\left| z \right|}^{2}}+2\left| z \right|-3}{\left| z \right|+1} \right)<-2 \\\ & \Rightarrow \dfrac{2{{\left| z \right|}^{2}}+2\left| z \right|-3}{\left| z \right|+1}>{{\left( \dfrac{1}{\sqrt{3}} \right)}^{-2}}=3 \\\ \end{aligned}$$ We multiply $\left| z \right|+1$ both side and have $$\begin{aligned} & \Rightarrow \dfrac{2{{\left| z \right|}^{2}}+2\left| z \right|-3}{\left| z \right|+1}>{{\left( \dfrac{1}{\sqrt{3}} \right)}^{-2}}=3 \\\ & \Rightarrow 2{{\left| z \right|}^{2}}+2\left| z \right|-3>3\left( \left| z \right|+1 \right) \\\ \end{aligned}$$ The inequality will not change as $\left| z \right|+1$ is positive. We subtract $3\left( \left| z \right|+1 \right)$ from both side and have $$\Rightarrow 2{{\left| z \right|}^{2}}-\left| z \right|-6>0$$ The left hand side of the above result is a quadratic equation. Let us substitute $t=\left| z \right|$ and putting above we break it by splitting the middle term method $$\begin{aligned} & \therefore 2{{t}^{2}}-t-6>0 \\\ & \Rightarrow 2{{t}^{2}}-4t+3t-6>0 \\\ & \Rightarrow 2t\left( t-2 \right)+3\left( t-2 \right)>0 \\\ & \Rightarrow \left( t-2 \right)\left( 2t+3 \right)>0 \\\ \end{aligned}$$ Let us put back $t=\left| z \right|$ in above result and get $$\left( \left| z \right|-2 \right)\left( 2\left| z \right|+3 \right)>0$$ We know that the product of two numbers is positive if and only if both the numbers are either positive or both the numbers are negative. We know that $\left| z \right|$ is always positive then $\left( 2\left| z \right|+3 \right)$ will always be positive. So other term $\left( \left| z \right|-2 \right)$ have to be positive. Now $$\begin{aligned} & \left( \left| z \right|-2 \right)>0 \\\ & \Rightarrow \left| z \right|>2 \\\ \end{aligned}$$ **So, the correct answer is “Option c”.** **Note:** We need to be careful of change in sign of logarithm inequality when the base lies between 0 and 1 because if the base is greater than 1 then the inequality sign will remain the same. If we express that in symbols, ${ {\log }_{a}} n < b \Rightarrow n < {{b}^{a}}$ when $ a > 1$.