If z= 3 +i , then the argument of z^2 e^ z-i is equal to

Step-by-step solution

Given z = 3 + i . Express z in polar form: z = r( + i ) , where r = |z| = ( 3 )^2 + 1^2 = 2 and = (z) = ⁻¹ ( 1 3 ) = 6 . Thus, z = 2e^ i /6 . We need to find the argument of w = z^2 e^ z-i . Taking the argument of both sides: (w) = (z^2 e^ z-i ) = (z^2) + (e^ z-i ) . Using properties of arguments: (z^2) = 2 (z) = 2 6 = 3 . For the second term, let z - i = x + iy . Since z = 3 + i , z - i = 3 + i - i = 3 . Therefore, e^ z-i = e^ 3 . Since e^ 3 is a positive real number, its argument is 0 . Thus, (w) = 3 + 0 = 3 . An