Correct Answer - A::B::D
Let z=-1-I and arg(z) = `theta`
Now `tan theta =|(lim (z))/(Re (z) )|=|(-1)/(-1)|=1`
`rArr theta =pi/4`
`Since x lt 0, y gt 0 `
`therefore (z) =- (pi-pi/4)=-(3pi)/(4)`
(b) We have ,f(t) =arg (-1 + it )
`arg(-1+it)={:(pi- tan^(-1) t"," t ge 0 ),-(pi + tan ^(-1))","t lt 0 ):}`
This function is discontinuous at t = 0.
(c ) We have
`arg (z_1/z_2)-arg (z_1)+arg(z_2)+2n pi `
`therefore arg(z_1/z_2)-arg (z_1)+arg(z_2)`
`therefore are (z_1/z_2)-arg (z_1)+arg (z_2)`
`= arg (z_1)-arg (z_2)+2 n pi -arg (z_1)+arg (z_2)`
`=2 n pi`
So given expression is multiple of `2 pi `
(d) We have arg `(((z-z_1)(z_2-z_3))/((z-z_3)(z_2-z_1)))=pi`
`rArr ((z-z_1)/(z-z_3))((z_2-z3)/(z_2-z_1))` is purely real
Thus , the points `A(z_1)B(z_2),C(z_3)` and D(z) taken in order would be concylic if purely real .
Hence, it is a circle
`threrefore (a),(b),(d)` are false statment.