If z satisfies `|z+1|lt|z-2|,` then `v=3z+2+i` satisfies:

Options

A. `|omega+1| lt |omega-8|`
B. `|omega+1| lt |omega-7|`
C. `omega+baromega gt 7`
D. `|omega +5| lt |omega-4|`

Answer

A

Explanation

We have,
`|z+1| lt |z-2|`
`rArr` z lies on the LHS of the perpendicular bisector of the line segment jioning (-1,0) and (2,0)
`rArr "Re"(z) lt 1/2`
`rArr 2"Re"(z) lt 1`
`rArr z+barz lt 1`
`rArr (omega-(2+i))/(3) +(bar(omega-(2+i))/(3) lt `
`|omega0(2+i) + baromega-(2-i) lt 3 rArr omega+baromega-4 lt 3 rArr omega+baromega lt 7`
Now, `|omega+1| lt |omega-8|`
`rArr omega` lies on the LHS of the perpendicular bisector of the line segment joining `(-1,0)` and (8,0).
`rArr "Re"(omega) lt 7/2 rArr 2"Re"(omega) lt 7 rArr omega+baromega lt 7`.
Hence, option (a) is correct.