If a complex number z satisfies $\log _{\frac{1}{2}}\left(\frac{|z|^{2}+2|z|+6}{2|z|^{2}-2|z|+1}\right)<0$, then locus of point represented by z is
Since $\log _{\frac{1}{2}}\left(\frac{|z|^{2}+2|z|+6}{2|z|^{2}-2|z|+1}\right)<0$
$\therefore \frac{|z|^{2}+2|z|+6}{2|z|^{2}-2|z|+1}<1$ $\Rightarrow|z|^{2}-4|z|-5>0$ $\Rightarrow(|z|-5)(|z|+1)>0 \Rightarrow|z|>5 \text{ and } |z|<-1$
But the ans is given in the question is $|z|<5$
Am i wrong???