Question

Solve the given inequality ${{\log }_{\dfrac{1}{3}}}\left[ \dfrac{{{\left| z-3 \right|}^{2}}+2}{11\left| z-3 \right|-2} \right]>1$, where z is a complex number.

Answer 1

We start solving the problem by assigning a variable for $\left| z-3 \right|$. We then use the facts ${{\log }_{\dfrac{1}{a}}}x=-{{\log }_{a}}x$, if ${{\log }_{a}}x < b$ $\left( a> 1 \right)$, then $x< {{a}^{b}}$ and ${{a}^{-1}}=\dfrac{1}{a}$ to proceed further through the problem. We then use the fact that if the inequality $\left( x-a \right)\left( x-b \right)< 0$, $\left( a< b \right)$ holds true, then the solution set for the variable x is $a < x < b$. We then assume the general form of the complex number for z and find the required solution.

Complete step-by-step solution:
According to the problem, we need to solve the inequality ${{\log }_{\dfrac{1}{3}}}\left[ \dfrac{{{\left| z-3 \right|}^{2}}+2}{11\left| z-3 \right|-2} \right]>1$.
Let us assume $\left| z-3 \right|=p$. So, we get ${{\log }_{\dfrac{1}{3}}}\left[ \dfrac{{{p}^{2}}+2}{11p-2} \right]>1$ ---(1).
We know that ${{\log }_{\dfrac{1}{a}}}x=-{{\log }_{a}}x$. We use this in equation (1).
$\Rightarrow -{{\log }_{3}}\left[ \dfrac{{{p}^{2}}+2}{11p-2} \right]>1$.
$\Rightarrow {{\log }_{3}}\left[ \dfrac{{{p}^{2}}+2}{11p-2} \right]<-1$ ---(2).
We know that if ${{\log }_{a}}x < b$ $\left( a > 1 \right)$, then $x<{{a}^{b}}$. We use this in equation (2).
$\Rightarrow \dfrac{{{p}^{2}}+2}{11p-2}<{{3}^{-1}}$ ---(3).
We know that ${{a}^{-1}}=\dfrac{1}{a}$. We use this in equation (3).
$\Rightarrow \dfrac{{{p}^{2}}+2}{11p-2}<\dfrac{1}{3}$.
$\Rightarrow 3\left( {{p}^{2}}+2 \right)<1\left( 11p-2 \right)$.
$\Rightarrow 3{{p}^{2}}+6<11p-2$.
$\Rightarrow 3{{p}^{2}}-11p+6+2<0$.
$\Rightarrow 3{{p}^{2}}-11p+8<0$.
$\Rightarrow 3{{p}^{2}}-3p-8p+8<0$.
$\Rightarrow 3p\left( p-1 \right)-8\left( p-1 \right)<0$.
$\Rightarrow \left( 3p-8 \right)\left( p-1 \right)<0$.
$\Rightarrow \left( p-\dfrac{8}{3} \right)\left( p-1 \right)<0$ ---(4).
We know that if the inequality $\left( x-a \right)\left( x-b \right)<0$, $\left( a < b \right)$ holds true, then the solution set for the variable x is $a < x < b$. We use this result for the equation (4).
So, the solution set for p for the inequality in equation (4) is $1 < p < \dfrac{8}{3}$. Now we substitute $p=\left| z-3 \right|$.
So, we get $1<\left| z-3 \right|<\dfrac{8}{3}$ ---(5).
Since z is a complex number, we assume $z=x+iy$. We substitute this in equation (5).
$\Rightarrow 1<\left| x+iy-3 \right|<\dfrac{8}{3}$.
$\Rightarrow 1<\left| \left( x-3 \right)+iy \right|<\dfrac{8}{3}$ ---(6).
We know that the magnitude of complex number $a+ib$ is defined as $\left| a+ib \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}$.
$\Rightarrow 1<\sqrt{{{\left( x-3 \right)}^{2}}+{{y}^{2}}}<\dfrac{8}{3}$ ---(7).
We know that if $a < x < b$ and $a>0$, $b>0$, then we get ${{a}^{2}} < {{x}^{2}} < {{b}^{2}}$. We use this result in equation (7).
$\Rightarrow {{1}^{2}} < {{\left( x-3 \right)}^{2}}+{{y}^{2}} < {{\left( \dfrac{8}{3} \right)}^{2}}$.
So, the solution set represents a circular ring or annulus with an inner radius of 1 unit and an outer radius of $\dfrac{8}{3}$ units.
$\therefore$ The solution for the given inequality ${{\log }_{\dfrac{1}{3}}}\left[ \dfrac{{{\left| z-3 \right|}^{2}}+2}{11\left| z-3 \right|-2} \right]>1$ is a circular ring or annulus with inner radius 1 units and outer radius $\dfrac{8}{3}$ units.

Note: We should not say that the solution set is the circles whose radius is between 1 and $\dfrac{8}{3}$. Because in that the region present inside radius 1 will also come as a solution for the inequality. We should not be confused with the formulas and make calculation mistakes while solving this problem. We can also find the solution set for x and y which is also an interval present between two real values. Whenever we see z in the problem, we take z as a complex number unless it is mentioned otherwise.