Let $ A = \left\{ \theta \in [0, 2\pi] : 1 + 10\operatorname{Re}\left( \frac{2\cos\theta + i\sin\theta}{\cos\theta - 3i\sin\theta} \right) = 0 \right\} $. Then $ \sum\limits_{\theta \in A} \theta^2 $ is equal to

If the locus of z ∈ ℂ, such that Re$ \left( \frac{z - 1}{2z + i} \right) + \text{Re} \left( \frac{\overline{z} - 1}{2\overline{z} - i} \right) = 2 $, is a circle of radius r and center $(a, b)$, then $ \frac{15ab}{r^2} $ is equal to :

Among the statements

(S1) : The set $\left\{z \in \mathbb{C}-\{-i\}:|z|=1\right.$ and $\frac{z-i}{z+i}$ is purely real $\}$ contains exactly two elements, and

(S2) : The set $\left\{z \in \mathbb{C}-\{-1\}:|z|=1\right.$ and $\frac{z-1}{z+1}$ is purely imaginary $\}$ contains infinitely many elements.

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