If $\log _{\frac{1}{\sqrt{3}}}\left\{\frac{|z|^2-|z|+1}{2+|z|}\right\}>-2$, then $\mathrm{z}$ lies inside

If $\log _{\frac{1}{\sqrt{3}}}\left\{\frac{|z|^2-|z|+1}{2+|z|}\right\}>-2$, then $\mathrm{z}$ lies inside
  1. a triangle
  2. an ellipse
  3. a circle
  4. a square

Solution

Given that, $ \log _{\frac{1}{\sqrt{3}}}\left\{\frac{|z|^2-|z|+1}{2+|z|}\right\}>-2 $ Since, $\log _a b>c, b < a^c$ if $0 < a < 1$ $ \begin{aligned} & \Rightarrow \quad \frac{|z|^2-|z|+1}{2+|z|} < \left(\frac{1}{\sqrt{3}}\right)^{-2} \\ & \Rightarrow \frac{|z|^2-|z|+1}{2+|z|} < (\sqrt{3})^2 \end{aligned} $ $ \begin{aligned} & \Rightarrow \quad|z|^2-|z|+1 < (\sqrt{3})^2(2+|z|) \\ & \Rightarrow \quad|z|^2-|z|+1 < 6+3|z| \\ & \Rightarrow \quad|z|^2-4|z|+1 < 6 \\ & \Rightarrow \quad(|z|-2)^2 < 9 \Rightarrow-3 < |z|-2 < 3 \\ & \Rightarrow \quad-1 < |z| < 5 \Rightarrow 0 < |z| < 5 \end{aligned} $ Hence, $\mathrm{z}$ lies inside the circle

Asked in: AP EAMCET 2017 (26 Apr Shift 1)

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