If `log_(1/3) |z+1| > log_(1/3) |z-1|` then prove that Re(z) < 0.

Options

A. `"Re"(z) ge 0`
B. `"Re"(z) lt 0`
C. `"Im"(z) gt 0`
D. None of these

Answer

B

Explanation

We have,
`log_(1//3)|z+1| gt log_(1//3)|z-1|`
`rArr |z+1| lt |z-1|` `[therefore log_(a)x gt log_(a)y rArr y rArr x lt 0` if `0 lt a lt 1]`
`rArr` Point representing z lies on the left side of the perpendicualr bisector of the line joining (-1,0) and (1,0)
`rArr` z lies on the left side of imaginary axis.
`rArr` Re(z) `lt0`