Question
If $${\log _{\frac{1}{2}}}\frac{{{{\left| z \right|}^2} + 2\left| z \right| + 4}}{{2{{\left| z \right|}^2} + 1}} < 0$$ then the region traced by $$z$$ is
A.
$$\left| z \right| < 3$$
B.
$$1 < \left| z \right| < 3$$
C.
$$\left| z \right| > 1$$
D.
$$\left| z \right| < 2$$
Answer :
$$\left| z \right| < 3$$
Solution :
$$\eqalign{
& {\log _{\frac{1}{2}}}\frac{{{{\left| z \right|}^2} + 2\left| z \right| + 4}}{{2{{\left| z \right|}^2} + 1}} < 0 = {\log _{\frac{1}{2}}}1 \cr
& \Rightarrow \,\,\frac{{{{\left| z \right|}^2} + 2\left| z \right| + 4}}{{2{{\left| z \right|}^2} + 1}} > 1 \cr
& {\text{or, }}{\left| z \right|^2} + 2\left| z \right| + 4 > 2{\left| z \right|^2} + 1 \cr
& {\text{or, }}{\left| z \right|^2} - 2\left| z \right| - 3 < 0 \cr
& {\text{or, }}\left( {\left| z \right| + 1} \right)\left( {\left| z \right| - 3} \right) < 0 \cr
& \therefore \,\,\left| z \right| - 3 < 0. \cr} $$