The solutions of the equation $z^2(1-z^2)=16$,$z \in \mathbb{C}$,lie on the curve

Let $S = \{z \in \mathbb{C} : |z-3| \leq 1 \text{ and } z(4+3i) + \bar{z}(4-3i) \leq 24\}$. If $\alpha + i\beta$ is the point in $S$ which is closest to $4i$,then $25(\alpha + \beta)$ is equal to

If $z=x+iy$ is a complex number satisfying $\left|\frac{z-2i}{z+2i}\right|=2$ and the locus of $z$ is a circle,then its radius is

If the eight vertices of a regular octagon are given by the complex numbers $z_j = \frac{1}{x_j - 2i}$ for $j = 1, 2, \dots, 8$,where $x_j$ are the roots of $x^8 - 1 = 0$,then the radius of the circumcircle of the octagon is

The equation $\arg \left(\frac{z-1}{z+1}\right)=\frac{\pi}{4}$ represents a circle with:

If $a = \operatorname{Im}\left(\frac{1+z^2}{2iz}\right)$ and $z$ is any non-zero complex number such that $|z|=1$,then $a=$