Let _1=(8+i)+(7+4i) and _2=(1+8i)+(4+7i).

Question

Let $\omega_1=(8+i)\sin	heta+(7+4i)\cos	heta$ and $\omega_2=(1+8i)\sin	heta+(4+7i)\cos	heta$. Let the product $\omega_1\omega_2=\alpha+i\beta$, where $i=\sqrt{-1}$. Let $p$ and $q$ be the maximum and the minimum values of $\alpha+\beta$ respectively. Then $p+q$ is equal to:

Accepted Answer

Write $\omega_1=a+ib,\ \omega_2=b+ia$ where $a=8\sin\theta+7\cos\theta,\ b=\sin\theta+4\cos\theta$.

$$\omega_1\omega_2=(a+ib)(b+ia)=ab+ia^2+ib^2-ab = i(a^2+b^2).$$

So $\alpha=0$ and $\beta=a^2+b^2$. Expanding:
$$a^2+b^2 = 65\sin^2\theta+65\cos^2\theta+(112+8)\sin\theta\cos\theta = 65+60\sin2\theta.$$

$$\alpha+\beta=65+60\sin2\theta, \quad p=(\alpha+\beta)_{\max}=125,\quad q=(\alpha+\beta)_{\min}=5.$$

$p+q=130.$

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