(3) If \( 8 x+y i=i^{-17}+24 \), where \( i^{2}=-1 \), then \( x-y=-1 \) \( \begin{array}{lll}\text { (a) } 2 & \text { (b) } 4 & \text { (c) }-4 \\ \text { (d) } 25\end{array} \)

First, let's simplify \( i^{-17} \). The powers of \( i \) cycle every four terms: - \( i^1 = i \) - \( i^2 = -1 \) - \( i^3 = -i \) - \( i^4 = 1 \) To find \( i^{-17} \), we first compute \( -17 \mod 4 \). This gives us \( -17 + 4 \cdot 5 = 3 \), so \( i^{-17} = i^3 = -i \). Now, substituting back into the original equation: \[ 8x + yi = -i + 24 \] This equation can be split into real and imaginary parts. The real part gives: \[ 8x = 24 \quad \Rightarrow \quad x = 3 \] The imaginary part gives: \[ y = -1 \] Now substituting \( x = 3 \) and \( y = -1 \) into \( x - y \): \[ x - y = 3 - (-1) = 3 + 1 = 4 \] Thus, the answer is: \[ \boxed{4} \]