Evaluate the following: \( \begin{array}{lllll}\text { a) } i^{13} & \text { (b) } 2 i^{3} & \text { (c) } i^{7} & \text { (d) } i^{44} & \text { (e) } i^{29}\end{array} \)

To evaluate the powers of \(i\), we can use the fact that \(i\) is the imaginary unit defined as \(i = \sqrt{-1}\), and it has a cyclic pattern based on its powers: - \(i^1 = i\) - \(i^2 = -1\) - \(i^3 = -i\) - \(i^4 = 1\) After \(i^4\), the powers repeat every four terms. Therefore, to find \(i^n\) for any integer \(n\), we can find \(n \mod 4\) and use the results above. Now, let's evaluate each case: **a) \(i^{13}\)**: Calculate \(13 \mod 4\): \[ 13 \div 4 = 3 \quad \text{(remainder 1)} \] So, \(i^{13} = i^1 = i\). **b) \(2i^{3}\)**: Calculate \(i^{3}\): \[ i^{3} = -i \] Thus, \(2i^{3} = 2(-i) = -2i\). **c) \(i^{7}\)**: Calculate \(7 \mod 4\): \[ 7 \div 4 = 1 \quad \text{(remainder 3)} \] So, \(i^{7} = i^3 = -i\). **d) \(i^{44}\)**: Calculate \(44 \mod 4\): \[ 44 \div 4 = 11 \quad \text{(remainder 0)} \] So, \(i^{44} = i^0 = 1\). **e) \(i^{29}\)**: Calculate \(29 \mod 4\): \[ 29 \div 4 = 7 \quad \text{(remainder 1)} \] So, \(i^{29} = i^1 = i\). Putting it all together: - \( \text{a) } i^{13} = i \) - \( \text{b) } 2i^{3} = -2i \) - \( \text{c) } i^{7} = -i \) - \( \text{d) } i^{44} = 1 \) - \( \text{e) } i^{29} = i \)