Complex Numbers And Quadratic Equations question 366
Question: If $ {\log_{\sqrt{3}}}( \frac{|z{{|}^{2}}-|z|+1}{2+|z|} ) $ $ <2 $ , then the locus of $ z $ is
Options:
A) $ |z|=5 $
B) $ |z|<5 $
C) $ |z|>5 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
We have $ {\log_{\sqrt{3}}}( \frac{|z{{|}^{2}}-|z|+1}{2+|z|} )<2 $
Þ $ \frac{|z{{|}^{2}}-|z|+1}{2+|z|}<{{(\sqrt{3})}^{2}} $
Þ $ |z{{|}^{2}}-4|z|-5<0 $
Þ $ -1<|z|<5 $
Þ $ |z|<5 $ as $ |z|>0 $ \ Locus of $ z $ is |z|<5
BETA
