Let $\left|\dfrac{\bar z-i}{2\bar z+i}\right|=\dfrac{1}{3},\ z\in\mathbb{C}$, 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:
Let $\left|\dfrac{z-i}{2z+i}\right|=\dfrac{1}{3}$, $z\in\mathbb{C}$, be the equation of a circle with centre $C$. If the area of the triangle whose vertices are at $(0,0)$, $C$, and $(\alpha,0)$ is 11 square units, then $\alpha^2$ equals
Let $z_1$, $z_2$, $z_3$ be three complex numbers lying on the circle $|z|=1$ with $\arg(z_1)=-\pi/4$, $\arg(z_2)=0$, $\arg(z_3)=\pi/4$. If $|z_1\bar{z}_2+z_2\bar{z}_3+z_3\bar{z}_1|^2 = \alpha+\beta\sqrt{2}$ where $\alpha,\beta\in\mathbb{Z}$, then $\alpha^2+\beta^2$ equals