Mathematical Derivation

Let the roots of the quadratic equation $Z^2 + pZ + q = 0$ be $Z_1$ and $Z_2$.

From Vieta's formulas, we have:

• $Z_1 + Z_2 = -p$
• $Z_1 Z_2 = q$

The points $A$ and $B$ represent $Z_1$ and $Z_2$ respectively in the complex plane, with $O$ denoting the origin.

The condition $OA = OB$ implies that the magnitudes of the roots are equal. Let $|Z_1| = |Z_2| = r$.

We can represent the roots in polar form as $Z_1 = r e^{i\theta_1}$ and $Z_2 = r e^{i\theta_2}$.

The condition $\angle AOB = \alpha \neq 0$ means the angle between the position vectors $OA$ and $OB$ is $\alpha$. This implies the difference between their arguments is $\alpha$. We can set $\theta_2 = \theta_1 + \alpha$.

Now, we express $p$ and $q$ using these root forms:

• $q = Z_1 Z_2 = (r e^{i\theta_1})(r e^{i\theta_2}) = r^2 e^{i(\theta_1+\theta_2)}$
• $p = -(Z_1 + Z_2) = -r(e^{i\theta_1} + e^{i\theta_2})$

Let's calculate the term $\frac{p^2}{q}$:

$p^2 = (-r(e^{i\theta_1} + e^{i\theta_2}))^2 = r^2 (e^{i\theta_1} + e^{i\theta_2})^2$

$p^2 = r^2 (e^{i2\theta_1} + 2e^{i\theta_1}e^{i\theta_2} + e^{i2\theta_2})$

Dividing $p^2$ by $q$:

$\frac{p^2}{q} = \frac{r^2 (e^{i2\theta_1} + 2e^{i(\theta_1+\theta_2)} + e^{i2\theta_2})}{r^2 e^{i(\theta_1+\theta_2)}}$

Simplify by dividing each term in the numerator by the denominator:

$\frac{p^2}{q} = \frac{e^{i2\theta_1}}{e^{i(\theta_1+\theta_2)}} + \frac{2e^{i(\theta_1+\theta_2)}}{e^{i(\theta_1+\theta_2)}} + \frac{e^{i2\theta_2}}{e^{i(\theta_1+\theta_2)}}$

$\frac{p^2}{q} = e^{i(\theta_1-\theta_2)} + 2 + e^{i(\theta_2-\theta_1)}$

Since $\theta_2 - \theta_1 = \alpha$, it follows that $\theta_1 - \theta_2 = -\alpha$. Substituting this into the expression:

$\frac{p^2}{q} = e^{-i\alpha} + 2 + e^{i\alpha}$

Using the identity $e^{ix} + e^{-ix} = 2\cos(x)$:

$\frac{p^2}{q} = (e^{i\alpha} + e^{-i\alpha}) + 2 = 2\cos(\alpha) + 2$

Now, apply the trigonometric identity $\cos(\alpha) = 2\cos^2(\frac{\alpha}{2}) - 1$:

$\frac{p^2}{q} = 2(2\cos^2(\frac{\alpha}{2}) - 1) + 2 = 4\cos^2(\frac{\alpha}{2}) - 2 + 2 = 4\cos^2(\frac{\alpha}{2})$

Final Calculation

We need to determine the value of the expression $\frac{p^2}{q} \sec^2 \frac{\alpha}{2}$.

Substitute the calculated value of $\frac{p^2}{q}$:

Value = $\left( 4\cos^2(\frac{\alpha}{2}) \right) \sec^2(\frac{\alpha}{2})$

Using the definition $\sec(x) = \frac{1}{\cos(x)}$, we have $\sec^2(\frac{\alpha}{2}) = \frac{1}{\cos^2(\frac{\alpha}{2})}$:

Value = $4\cos^2(\frac{\alpha}{2}) \times \frac{1}{\cos^2(\frac{\alpha}{2})}$

The $\cos^2(\frac{\alpha}{2})$ terms cancel out, provided $\cos(\frac{\alpha}{2}) \neq 0$ (which is true since $\alpha \neq 0$ and $\alpha \neq 2\pi + 2k\pi$ for integer $k$ leading to $\alpha/2 = \pi/2 + k\pi$):

Value = $4$

The value of the expression $\frac{p^2}{q} \sec^2 \frac{\alpha}{2}$ is 4.