The image contains two questions related to the Z-transform of given sequences, which appear to involve sinusoidal and hyperbolic trigonometric functions.

Let's break down each question.

7) The Z-transform of the sequence $u_n = \sin(2n - 3)$ is:

The Z-transform for a sinusoidal sequence like $\sin(\omega n + \theta)$ can be calculated using the standard result:

$Z\{ \sin(\omega n + \theta) \} = \frac{z \sin \theta - z^2 \sin (\omega + \theta)}{z^2 - 2z \cos(\omega) + 1}, \quad |z| > 1$

In this case, we have $\omega = 2$ and $\theta = -3$. We need to apply the formula, and based on the choices provided, we can evaluate the correct result by calculating the numerator and denominator step by step.

8) The Z-transform of $\cosh \left( n\frac{\pi}{2} + \theta \right)$:

For hyperbolic functions like $\cosh(\omega n + \theta)$, the process is similar to sinusoidal functions, but instead of sine and cosine, the hyperbolic counterparts $\sinh$ and $\cosh$ are involved. The Z-transform can be given by:

$Z\{ \cosh(\omega n + \theta) \} = \frac{z \cosh(\theta) - \cosh(\omega)}{z^2 - 2 z \cosh(\omega) + 1}, \quad |z| > |e^{\omega/2}|$

Here, $\omega = \frac{\pi}{2}$, and we can substitute that value into the general formula to simplify it, yielding one of the provided options.

Would you like me to calculate the exact Z-transforms for each of these, or do you need help with a specific step in the process? Let me know if you'd like details!

5 Follow-up questions:

1. What are the steps to compute the Z-transform of any sinusoidal function?
2. How does the Z-transform handle hyperbolic functions like $\cosh$ and $\sinh$?
3. Why does the region of convergence $|z| > 1$ or $|z| > e^{\pi/2}$ appear in these problems?
4. How do you differentiate between the numerators for sine and cosine Z-transform formulas?
5. What changes when you compute the Z-transform for sequences involving phase shifts?

Tip: The Z-transform is especially useful for solving difference equations and analyzing discrete-time systems in signal processing.