# A question based on complex number

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.

• Are you able to solve this numerically? Because that logarithm looks intimidating – Alex S Nov 23 '18 at 12:17

Since $$\cos {\pi\over6}=\frac{\sqrt 3}{2}<1,$$ 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 $$\frac{|z–2|+5}{4|z–2|–4}>{3\over 4}$$ 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: $$|z-2|+5>3|z-2|-3\implies |z-2|<4$$ 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).

• We need to expand it further – priyanka kumari Nov 23 '18 at 15:30