If $z_1$ and $z_2$ are two roots of the equation $z^2+az+b=0$ with $a^2 < 4b$,then the origin,$z_1$,and $z_2$ form an equilateral triangle if

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

If $|z_1|=1, |z_2|=2, |z_3|=3$ and $|9z_1z_2 + 4z_1z_3 + z_2z_3| = 12$,then the value of $|z_1 + z_2 + z_3|$ is equal to :-

If $|z + 4| \le 3$,then the greatest and the least value of $|z + 1|$ are

The points in the set $\{z \in \mathbb{C} : \arg \left(\frac{z-2}{z-6i}\right) = \frac{\pi}{2}\}$ (where $\mathbb{C}$ denotes the set of all complex numbers) lie on the curve which is a

If $z$ is a complex number such that $|z| \ge 2$,then the minimum value of $|z + \frac{1}{2}|$ is: