Concept of Log

Question

!!log_(tan (pi/6)) ((|z|^2-|z|+1)/(2+|z|))> -2!!,

For any !!z!! which satisfy the above equation !!|z|!! is less than ?

(a) 6

(b) 5

(c) 4

(d) 3

Solution

!!log_(tan (pi/6)) ((|z|^2-|z|+1)/(2+|z|))> -2!!

Since !!0< tan(pi/6)<1!!,inequality will reverse.

Hence,

!!(|z|^2-|z|+1)/(2+|z|) < (1/sqrt(3))^-2 = (sqrt3)^2 = 3 !!

!!=>(|z|^2-|z|+1)/(2+|z|) < 3!!

!!=> |z|^2-|z|+1 < 3 (2+|z|)!!

!!=> |z|^2-4|z|-5 < 0!!

!!=> (|z|-5)(|z|+1)< 0!!

!!=> |z| in (-1,5)!!

Since !!|z|!! can't be negative, therefore !!|z| in [0,5)!!

Option (b) is correct.