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The question is $$\log_{\frac{1}{√3}}\frac{|z|^2–|z|+1}{2+|z|}>–2.$$ I tried to solve this and simplified it to $$|z|^2-4|z|-5<0.$$

But I wasn't able to locate this number, please tell me the procedure as well as the answer how to do it without actually plotting it on the graph as use of graph is not allowed. Assume $z=x+iy$.

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    $\begingroup$ use polar coordinates. In polar coordinates we have $z=|z| e^{i\alpha}$ for some $\alpha\in(-\pi, \pi]$. Then, from your inequality, you can estimate $r$. However the inequality doesn't defines a unique possible solution $\endgroup$ – Masacroso Nov 27 '18 at 3:09
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Hint: $|z|\ge 0$, and $$|z|^2-4|z|-5 = (|z|-2)^2-9<0 \implies -3<|z|-2<3 \implies ?$$

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  • $\begingroup$ I got some hint it will be a circle with radius 2 units but what after that $\endgroup$ – priyanka kumari Nov 27 '18 at 4:49
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$|z|$ is the distance from $0$ to the point $z,$ denote it $r.$
Then $|z|^2-4|z|-5<0$ rewrites $r^2-4r-5<0$ and is equivalent to $$(r-5)(r+1)<0.$$ Since $(r+1)$ is clearly positive (it is a distance + $1$), we solve $$r-5<0$$ or, in terms of $z,$ $$|z|<5$$

Geometrically: the solutions form the circle centered in $0$, with radius $5,$ without the boundary circumference.

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  • $\begingroup$ Thanks for your help $\endgroup$ – priyanka kumari Nov 30 '18 at 0:58

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