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

Let's evaluate each expression using the properties of the imaginary unit \( i \), where \( i = \sqrt{-1} \) and the cycle of powers of \( i \) repeats every four: \[ \begin{align*} i^1 &= i \\ i^2 &= -1 \\ i^3 &= -i \\ i^4 &= 1 \\ i^5 &= i \quad (\text{and so on}) \end{align*} \] For any integer exponent \( n \), the value of \( i^n \) depends on the remainder when \( n \) is divided by 4. Let's apply this to each part: ### a) \( i^{13} \) \[ 13 \div 4 = 3 \text{ with a remainder of } 1 \\ \Rightarrow i^{13} = i^1 = i \] ### b) \( 2i^{3} \) \[ 3 \div 4 = 0 \text{ with a remainder of } 3 \\ \Rightarrow i^{3} = -i \\ \Rightarrow 2i^{3} = 2(-i) = -2i \] ### c) \( i^{7} \) \[ 7 \div 4 = 1 \text{ with a remainder of } 3 \\ \Rightarrow i^{7} = i^{4 \times 1 + 3} = i^4 \cdot i^3 = 1 \cdot (-i) = -i \] ### d) \( i^{44} \) \[ 44 \div 4 = 11 \text{ with a remainder of } 0 \\ \Rightarrow i^{44} = i^{4 \times 11} = (i^4)^{11} = 1^{11} = 1 \] ### e) \( i^{29} \) \[ 29 \div 4 = 7 \text{ with a remainder of } 1 \\ \Rightarrow i^{29} = i^1 = i \] ### Summary of Results: \[ \begin{align*} \text{a) } i^{13} &= i \\ \text{b) } 2i^{3} &= -2i \\ \text{c) } i^{7} &= -i \\ \text{d) } i^{44} &= 1 \\ \text{e) } i^{29} &= i \end{align*} \] Answer: a) i; b) –2 i; c) –i; d) 1; e) i.