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:

Let the curve $z(1+i)+\bar{z}(1-i)=4, z \in C$, divide the region $|z-3| \leq 1$ into two parts of areas $\alpha$ and $\beta$. Then $|\alpha-\beta|$ equals :

Let $z_1, z_2$ and $z_3$ be three complex numbers on the circle $|z|=1$ with $\arg \left(z_1\right)=\frac{-\pi}{4}, \arg \left(z_2\right)=0$ and $\arg \left(z_3\right)=\frac{\pi}{4}$. If $\left|z_1 \bar{z}_2+z_2 \bar{z}_3+z_3 \bar{z}_1\right|^2=\alpha+\beta \sqrt{2}, \alpha, \beta \in Z$, then the value of $\alpha^2+\beta^2$ is :

Let $$z$$ be a complex number such that the real part of $$\frac{z-2 i}{z+2 i}$$ is zero. Then, the maximum value of $$|z-(6+8 i)|$$ is equal to