If $x^2+x+1=0$, then the value of $\left(x+\frac{1}{x}\right)^4+\left(x^2+\frac{1}{x^2}\right)^4+\left(x^3+\frac{1}{x^3}\right)^4+\ldots+\left(x^{25}+\frac{1}{x^{25}}\right)^4$ is:

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.