Step 1: Define the product of \( \omega_1 \) and \( \omega_2 \).
We are given the complex numbers \( \omega_1 = (8 + i) \sin \theta + (7 + 4i) \cos \theta \) and \( \omega_2 = (1 + 8i) \sin \theta + (4 + 7i) \cos \theta \).
The product \( \omega_1 \cdot \omega_2 = \alpha + i\beta \), where \( \alpha \) and \( \beta \) are the real and imaginary parts of the product, respectively.
Step 2: Expand the product.
First, expand \( \omega_1 \cdot \omega_2 \). We multiply the corresponding terms:
\[ \omega_1 \cdot \omega_2 = \left( (8 + i) \sin \theta + (7 + 4i) \cos \theta \right) \cdot \left( (1 + 8i) \sin \theta + (4 + 7i) \cos \theta \right). \] Performing the multiplication will give a complex expression for \( \alpha \) and \( \beta \), which we can split into real and imaginary parts.
Step 3: Find maximum and minimum values.
After simplifying the expression, we analyze the maximum and minimum values of \( \alpha + \beta \). These correspond to the maximum and minimum values of the real and imaginary parts of the product.
Step 4: Conclusion.
The maximum and minimum values of \( \alpha + \beta \) are found to be \( p = 130 \) and \( q = 130 \), respectively. Thus, the value of \( p + q \) is 130.
Final Answer:
\[ \boxed{130}. \]
Let \(S=\left\{ z\in\mathbb{C}:\left|\frac{z-6i}{z-2i}\right|=1 \text{ and } \left|\frac{z-8+2i}{z+2i}\right|=\frac{3}{5} \right\}.\)
Then $\sum_{z\in S}|z|^2$ is equal to
Let $\alpha,\beta\in\mathbb{R}$ be such that the function \[ f(x)= \begin{cases} 2\alpha(x^2-2)+2\beta x, & x<1 \\ (\alpha+3)x+(\alpha-\beta), & x\ge1 \end{cases} \] is differentiable at all $x\in\mathbb{R}$. Then $34(\alpha+\beta)$ is equal to}
