Let's tackle the problem step by step: --- ### **Step 1: Write the Z-transform Definition** The Z-transform of a discrete-time signal \( f(k) \) is defined as: \[ F(z) = \sum_{k=0}^{\infty} f(k) z^{-k} \] Given \( f(k) = \sin(\omega k) \), substitute this into the Z-transform definition: \[ F(z) = \sum_{k=0}^{\infty} \sin(\omega k) z^{-k} \] **Explanation:** We've set up the infinite sum for the Z-transform using the given \( f(k) \). --- ### **Step 2: Simplify the Sum** Recall that the sine function can be written in terms of exponentials: \[ \sin(\omega k) = \frac{e^{j\omega k} - e^{-j\omega k}}{2j} \] Substitute this into the sum: \[ F(z) = \sum_{k=0}^{\infty} \left[ \frac{e^{j\omega k} - e^{-j\omega k}}{2j} \right] z^{-k} = \frac{1}{2j} \sum_{k=0}^{\infty} \left[ (e^{j\omega} z^{-1})^k - (e^{-j\omega} z^{-1})^k \right] \] Both terms are geometric series, so use the formula for the sum of an infinite geometric series, \( \sum_{k=0}^{\infty} r^k = \frac{1}{1 - r} \) for \( |r| < 1 \): \[ F(z) = \frac{1}{2j} \left[ \frac{1}{1 - e^{j\omega}z^{-1}} - \frac{1}{1 - e^{-j\omega}z^{-1}} \right] \] **Explanation:** We used the exponential form of sine and the geometric series formula to simplify the Z-transform. --- ## **Final Answer:** \[ \boxed{ F(z) = \frac{1}{2j} \left[ \frac{1}{1 - e^{j\omega}z^{-1}} - \frac{1}{1 - e^{-j\omega}z^{-1}} \right] } \] This is the Z-transform of \( f(k) = \sin(\omega k) \) using the infinite sum definition.