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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.

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

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$$\log_{\cos(\pi/6)} \frac{|z–2|+5}{4|z–2|–4}<2$$ 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).

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  • $\begingroup$ We need to expand it further $\endgroup$
    – user567571
    Nov 23 '18 at 15:30

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