The shaded region, where `P=(-1,0),Q=(-1+sqrt(2),sqrt(2))R=(-1+sqrt(2),-sqrt(2),S=(1,0)` is represented by Figure `|z+1|>2,|a r g(z+1)2,|a r g(z+1)>>pi/4` `|z+1|<<2,|a r g(z+1)>pi/2`

Options

A. `|z+1| gt 2, |"arg"(z+1)| lt pi/4`
B. `|z+1| lt 2,|"arg"(z+1)| lt pi/4`
C. `|z-1| gt 2, |"arg"(z+1)| gt pi/4`
D. `|z-1|lt 2, |"arg"(z+1)| gt pi/2`

Answer

A

Explanation

We have,
`PQ=PR=PS`
`therefore` QR is an arc of the circle centred at `P(-1,0)` and radius 2 units.
Also, `angleQPS=angleRPS=pi/4`
Therefore, any point z in the shaded region lies outside the circle `|z+1|=2` and the line joining z and `P(-1,0)` makes angle less than `45^(@)` with X-axis on its either side.
Hence, `|z+1| gt 2` and `|"arg"(z+1)| lt pi/4`.