Question

If `arg(z) < 0` then `arg(-z)-arg(z)=` (1) `pi` (2) `-pi` (3) `- pi/2` (4) `pi/2`
A. `pi`
B. `-pi`
C. `- pi//2`
D. `pi //2`

Answer 1

Correct Answer - A
Since arg (z) `lt 0 rArr arg (z) =-theta`

`rArr z=r cos (- theta )`
`=r (cos theta - I sin theta)`
`-z = -r [ cos (pi - theta)+ I sin (pi - theta )]`
`therefore arg (-z) = pi - theta`
Thus ,arg (-z)-arg (z)
`= pi - theta - (- theta)=pi`
Alternate Solution
Reason arg (-z)-arg `((-z)/(z)) = arg (-1) =pi `