\$. a) The value of \( i^{24} \) is -.... b) Find the multiplicative inverse of \( 1- \) in atib form

To solve the problems, let's break them down one by one. ### a) The value of \( i^{24} \) The imaginary unit \( i \) is defined as \( i = \sqrt{-1} \). The powers of \( i \) cycle every four terms: - \( i^1 = i \) - \( i^2 = -1 \) - \( i^3 = -i \) - \( i^4 = 1 \) After \( i^4 \), the cycle repeats: - \( i^5 = i \) - \( i^6 = -1 \) - \( i^7 = -i \) - \( i^8 = 1 \) To find \( i^{24} \), we can determine the equivalent exponent modulo 4: \[ 24 \mod 4 = 0 \] Since \( 24 \equiv 0 \mod 4 \), we have: \[ i^{24} = i^0 = 1 \] So, the value of \( i^{24} \) is \( 1 \). ### b) Find the multiplicative inverse of \( 1 - i \) The multiplicative inverse of a complex number \( z = a + bi \) is given by: \[ \frac{1}{z} = \frac{\overline{z}}{|z|^2} \] where \( \overline{z} \) is the complex conjugate of \( z \) and \( |z|^2 = a^2 + b^2 \). For \( z = 1 - i \): 1. The complex conjugate \( \overline{z} = 1 + i \). 2. The modulus squared \( |z|^2 = (1)^2 + (-1)^2 = 1 + 1 = 2 \). Now, we can find the multiplicative inverse: \[ \frac{1}{1 - i} = \frac{1 + i}{2} \] Thus, the multiplicative inverse of \( 1 - i \) in standard form is: \[ \frac{1}{2} + \frac{1}{2}i \] In summary: a) \( i^{24} = 1 \) b) The multiplicative inverse of \( 1 - i \) is \( \frac{1}{2} + \frac{1}{2}i \).