Ytukyg

Locate the complex number $z=x+iy;$ for which $$log_{cos(pi/6)} frac{|z–2|+5}{4|z–2|–4}<2$$ I tried to solve this problem but is equation of circle and then putting the values but I was not able to proceed further please help me out.

Since $cos {piover6}=frac{sqrt 3}{2}<1,$ the logarithme is decreasing.

The number $|z-2|$ is the distance between $2$ and $z.$ For simplicity, denote $|z-2|=a ;$ and solve the equivalent inequality * $$quad frac{a+5}{4(a-1)}>{frac 34} tag 1$$
or $$frac{2(4-a)}{a-1}>0. tag 2$$

The solutions are $a in (1,4)$ or, in terms of $z,$ $$1<|z-2|<4.$$ Convenient points fulfill the open area bounded by two concentric circles with center $z_0=2,$ radii $r=1$ and $R=4, $ respectively.

*Note: It is not necessary to care about the domain of the function, as due to $(1)$ is the logarithme well defined.

$$log_{cos(pi/6)} frac{|z–2|+5}{4|z–2|–4}<2$$ so we have

$$ frac{|z–2|+5}{4|z–2|–4}>cos^2(pi/6)$$

so $$ frac{|z–2|+5}{4|z–2|–4}>{3over 4}$$ Since the right side is positive then the left side must be also. But denumerator is always positive so numerator must be also, so we have $|z-2|>1$. Now we can also get rid of the fractions and we get: $$|z-2|+5>3|z-2|-3implies |z-2|<4$$
If we put this together we get $$1<|z-2|<4$$
so $z$ is betwen two concentric circles with radius 1 and 4 and with center at $4$ (this is a point on real axsis).