The minimum value of $|z-1|+|z-5|$ is

Let $z_{1}$ and $z_{2}$ be the roots of the equation $z^{2} + az + 12 = 0$. If $z_{1}$,$z_{2}$,and the origin form an equilateral triangle in the complex plane,then the value of $|a|$ is:

The polar coordinates of the point,whose Cartesian coordinates are $(-2 \sqrt{3}, 2)$,are

$\omega$ is a complex cube root of unity and $Z$ is a complex number satisfying $|Z-1| \leq 2$. The possible values of $r$ such that $|Z-1| \leq 2$ and $|\omega Z - 1 - \omega^2| = r$ have no common solution are

For any integer $k$,let $\alpha_k = \cos \left(\frac{k \pi}{7}\right) + i \sin \left(\frac{k \pi}{7}\right)$,where $i = \sqrt{-1}$. The value of the expression $\frac{\sum_{k=1}^{12} |\alpha_{k+1} - \alpha_k|}{\sum_{k=1}^3 |\alpha_{4k-1} - \alpha_{4k-2}|}$ is

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