<s>[INST] Solving complex number by geometrical method. Let C1
 and C2
 are concentric circles of radius $1$ and $8/3$ respectively having centre at $(3, 0)$ on the argand
plane. If the complex number $z$ satisfies the inequality, $\log_{1/3}{\frac
{|z-3|^2+2}{11|z-3 |-2}}> 1$ then :
(A) $z$ lies outside C1
 but inside C2
(B) $z$ lies inside of both C1
 and C2
(C) $z$ lies outside both of C1
 and C2
(D) none of these
I tried to do it like this.${\frac
{|z-3|^2+2}{11|z-3 |-2}}<\frac{1}{3}$.So by simplifying I got ${\frac
{3|z-3|^2+8-11|z-3 |}{11|z-3 |-2}}<0$ .Hence by substituting $|z-3|$ by $t$ I get $t=1$ and $t=8/3$.So I get the negative value between $1$ and $8/3$.But I don't know how to proceed and check which option is correct.
 [/INST] $z$ is in the annulus formed by the two circles. Answer (A).
Indeed set $p=|z-3|$
$\log_{1/3}{\dfrac
{p^2+2}{11p-2}}> 1$ 
is verified if 
$0<\dfrac
{p^2+2}{11p-2}<\dfrac{1}{3}$
Let's solve
$\dfrac
{p^2+2}{11p-2}>0$ numerator is always positive so must be $11p-2>0\to p>\dfrac{2}{11}$
and 
$\dfrac
{p^2+2}{11p-2}<\dfrac{1}{3}\to \dfrac
{p^2+2}{11p-2}-\dfrac{1}{3}<0 \to \dfrac{3p^2+6-11p+2}{3(11p-2)}<0\to \dfrac{3p^2-11p+8}{3(11p-2)}<0$
As the denominator must be positive for the previous condition, then the numerator must be negative, that is
$3p^2-11p+8<0\to 1<p<\dfrac{8}{3}$
Thus $1<|z-3|<\dfrac{8}{3}$
which means that $z$ is outside $C_1$ but inside $C_2$ as stated above.
Hope this helps 
</s>