Let $O$ be the origin, the point $A$ be $z_1=\sqrt{3}+2 \sqrt{2} i$, the point $B\left(z_2\right)$ be such that $\sqrt{3}\left|z_2\right|=\left|z_1\right|$ and $\arg \left(z_2\right)=\arg \left(z_1\right)+\frac{\pi}{6}$. Then

If $\alpha$ and $\beta$ are the roots of the equation $2 z^2-3 z-2 i=0$, where $i=\sqrt{-1}$, then $16 \cdot \operatorname{Re}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right) \cdot \operatorname{lm}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)$ is equal to

The number of complex numbers $z$, satisfying $|z|=1$ and $\left|\frac{z}{\bar{z}}+\frac{\bar{z}}{z}\right|=1$, is :

Let $\left|\frac{\bar{z}-i}{2 \bar{z}+i}\right|=\frac{1}{3}, z \in C$, be the equation of a circle with center at $C$. If the area of the triangle, whose vertices are at the points $(0,0), C$ and $(\alpha, 0)$ is 11 square units, then $\alpha^2$ equals: