Find the pricipal argument of each of the following:
(a) `-1-isqrt(3)`
(b) `(1+sqrt(3)i)/(3 +i)`
(c) `sin alpha + i(1-cos alpha), 0 gt alpha gt pi`
(d) `(1+ isqrt(3))^(2)`

Explanation

(a) Let `z = - 1 - isqrt(3)`. Then `alpha = tan^(-1)|(b)/(a)| = tan^(-1)|(sqrt(3))/(1)| = (pi)/(3)`
Clearly, z lies in III quadrant.
Therefore, argument `theta = - (pi-alpha) = - (pi- pi//3) = (-2pi)/(3)`
(b) `z = (1+sqrt(3i))/(sqrt(3) +i) = ((1+sqrt(3i))(sqrt(3)-i))/((sqrt(3) + i)(sqrt(3) + i))`
`= (2sqrt(3) + 2i)/(4) = (sqrt(3) + i)/(2)`
`agr(z) = tan^(-1). (1)/(sqrt(3))= (pi)/(6)`
(c) `z = sin alpha + i (1-cos alpha)`
`rArr arg(z)= tan^(-1)((1-cos alpha)/(sin alpha)) = tan^(-1) ((2sin^(2)(alpha)/(2))/(2sin.(alpha)/(2)cos.(alpha)/(2)))`
`= tan^(-1) tan((alpha)/(2)) =(alpha)/(2)`
(d) `z = (1+isqrt(3))^(2) = 1-3 + 2isqrt(3) = - 2+ i(2sqrt(3))`
So, z lies in second quadrant.
`therefore arg(z) = pi - tan^(-1)|(2sqrt(3))/(-2)|`
`= pi - tan^(-1) sqrt(3) = pi - (pi)/(3) = (2pi)/(3)`