What is the argument of the complex number (1+i)(2+i) 3-i , where i= -1 ?

Step-by-step solution

Let z = (1+i)(2+i) 3-i . First, simplify the numerator: (1+i)(2+i) = 2 + i + 2i + i² = 2 + 3i - 1 = 1 + 3i . Now, multiply the numerator and denominator by the conjugate of the denominator (3+i) : z = (1+3i)(3+i) (3-i)(3+i) = 3 + i + 9i + 3i² 3² + 1² = 3 + 10i - 3 9 + 1 = 10i 10 = i . The complex number z = i can be written in polar form as 0 + 1i . The argument of z = i is arg (i) = 2 . Answer: 2