Product of Complex Numbers Calculation

Let the given complex numbers be:

$\omega_1 = (8 + i)\sin\theta + (7+4i) \cos\theta$

$\omega_2 = (1 + 8i)\sin\theta + (4+7i) \cos\theta$

We can rewrite them by grouping terms with $\sin\theta$ and $\cos\theta$:

$\omega_1 = (8\sin\theta + 7\cos\theta) + i(\sin\theta + 4\cos\theta)$

$\omega_2 = (\sin\theta + 4\cos\theta) + i(8\sin\theta + 7\cos\theta)$

Let $A = 8\sin\theta + 7\cos\theta$ and $B = \sin\theta + 4\cos\theta$. Then, $\omega_1 = A + iB$ and $\omega_2 = B + iA$.

Finding the Product $\alpha + i\beta$

The product $\omega_1 \times \omega_2$ is:

$\omega_1 \times \omega_2 = (A + iB)(B + iA)$

$= AB + iA^2 + iB^2 + i^2 AB$

$= AB + i(A^2 + B^2) - AB$

$= i(A^2 + B^2)$

Comparing this with $\alpha + i\beta$, we get:

$\alpha = 0$

$\beta = A^2 + B^2$

Calculating $\beta = A^2 + B^2$

Now, we calculate $A^2$ and $B^2$:

$A^2 = (8\sin\theta + 7\cos\theta)^2 = 64\sin^2\theta + 112\sin\theta\cos\theta + 49\cos^2\theta$

$B^2 = (\sin\theta + 4\cos\theta)^2 = \sin^2\theta + 8\sin\theta\cos\theta + 16\cos^2\theta$

Adding them:

$A^2 + B^2 = (64+1)\sin^2\theta + (112+8)\sin\theta\cos\theta + (49+16)\cos^2\theta$

$= 65\sin^2\theta + 120\sin\theta\cos\theta + 65\cos^2\theta$

Using $\sin^2\theta + \cos^2\theta = 1$ and $2\sin\theta\cos\theta = \sin(2\theta)$:

$A^2 + B^2 = 65(\sin^2\theta + \cos^2\theta) + 60(2\sin\theta\cos\theta)$

$= 65(1) + 60\sin(2\theta)$

$= 65 + 60\sin(2\theta)$

So, $\beta = 65 + 60\sin(2\theta)$.

Finding Maximum ($p$) and Minimum ($q$) Values

We need to find the maximum and minimum values of $\alpha + \beta$. Since $\alpha = 0$, we need the extrema of $\beta$.

$\alpha + \beta = 0 + (65 + 60\sin(2\theta)) = 65 + 60\sin(2\theta)$

The range of $\sin(2\theta)$ is $[-1, 1]$.

• Maximum value ($p$): Occurs when $\sin(2\theta) = 1$. $p = 65 + 60(1) = 125$.
• Minimum value ($q$): Occurs when $\sin(2\theta) = -1$. $q = 65 + 60(-1) = 65 - 60 = 5$.

Calculating $p + q$

Finally, calculate the sum $p + q$:

$p + q = 125 + 5 = 130$.