First slide

Algebra of complex numbers

Question

$If \log_{1 / \sqrt{3}} \frac{| z |^{2} - | z | + 1}{2 + | z |} > - 2, then$

Moderate

A

$|z| = 5$
B

$|z| < 5$
C

$|z| > 5$
D

$|z| > 10$

Solution

$Since \log_{\frac{1}{\sqrt{3}}} \frac{| z |^{2} - | z | + 1}{2 + | z |} > - 2$
$= \log_{\frac{1}{\sqrt{3}}} {(\frac{1}{\sqrt{3}})}^{- 2}$
$\Rightarrow \frac{| z |^{2} - | z | + 1}{2 + | z |} < {(\frac{1}{\sqrt{3}})}^{- 2}$ $[∵ \log_{a} x > \log_{a} y \Rightarrow x < y if 0 < a < 1]$ :

$\Rightarrow {|z|}^{2} - |z| + 1 < 6 + 3 |z|$
$\Rightarrow {|z|}^{2} - 4 |z| - 5 < 0$
$\Rightarrow (|z| - 5) (|z| + 1) < 0$
$But | z | + 1 > 0$
$∴ |z| - 5 < 0$
$\Rightarrow |z| < 5$

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