For a non-zero complex number `z` , let `arg(z)` denote the principal argument with `pi lt arg(z)leq pi` Then, which of the following statement(s) is (are) FALSE? `arg(-1,-i)=pi/4,` where `i=sqrt(-1)` (b) The function `f: R->(-pi, pi],` defined by `f(t)=arg(-1+it)` for all `t in R` , is continuous at all points of `RR` , where `i=sqrt(-1)` (c) For any two non-zero complex numbers `z_1` and `z_2` , `arg((z_1)/(z_2))-arg(z_1)+arg(z_2)` is an integer multiple of `2pi` (d) For any three given distinct complex numbers `z_1` , `z_2` and `z_3` , the locus of the point `z` satisfying the condition `arg(((z-z_1)(z_2-z_3))/((z-z_3)(z_2-z_1)))=pi` , lies on a straight line

Options

A. `arg (-1 -i) = (pi)/(4),` where `i = sqrt(-1)`
B. The function` f: R to (-pi,pi]`, defined by `f(t) = arg (-1 + it)` for all ` t in R`, is continous at all points of R, where `i=sqrt(-1)`
C. For any tow non-zero complex number `z_(1)` and `z_(2) , arg((z_(1))/(z_(2)) -arg(z_(1)) + arg(z_(2))` is an integer multiple of `2pi`
D. For any three given distinct complex numbers `z_(1),z_(2) and z_(3)` the locus of the point z satisfying the condition `(((z-z_(1))(z_(2)-z_(3)))/((z-z_(3))(z_(2)-z_(1)))) = pi`, lies on a strainght line.

Answer

A, B, D

Explanation

(1) Comple number -1- I lies in thrid quadrant
`therefore arg (-1-i) = -(3pi)/(4)`
(2) We have `f:R to (-pi,pi],f(t) =arg(-1+it)` for all `t in R`
Complex number `-1 + it` lies in second quadrant if `tgt 0` and in third quadrant if `t lt 0`.
`f(t) =arg (-1 +it)`
`{{:(,pi-tan^(-1)|(t)/(-1)|, tge0),(,-(pi-tan^(-)|(t)/(-1)|),t lt0):}`
`{{:(, pi-tan^(-1) t, t ge 0),(,-pi + tan^(-1) t, t le 0):}`
Clearly f(x) is discountinous at = 0.
(3) `arg((z_(1))/(z_(2))) -arg (z_(1)) + arg(z_(2))`
`= arg(z_(1)) -argz_(2) + 2npi -argz_(1)+arg z_(2) = 2npi, n in I`
(4) ` arg(((z-z_(1))(z_(2)-z_(3)))/((z-z_(3))(z_(2)-z_(1)))) = pi`
`rArr arg.(z-z_(1))/(z-z_(3))+ arg.(z_(2)-z_(3))/(z_(2_(beta)) - z_(1)) = pi`
when `alpha` and `beta` are angles subtened by line segment joining `z_(1)` and `z_(3)` at z and `z_(2)` respectively.
So, we get the following figure.
Thus, locus of z is a circle .