For any complex number $z,\bar z = \left( {\frac{1}{z}} \right)$if and only if
$z$ is a pure real number
$|z| = 1$
$z$ is a pure imaginary number
$z = 1$
$\left| {\frac{1}{2}({z_1} + {z_2}) + \sqrt {{z_1}{z_2}} } \right| + \left| {\frac{1}{2}({z_1} + {z_2}) - \sqrt {{z_1}{z_2}} } \right|$ =
Let $\alpha=8-14 i , A=\left\{ z \in C : \frac{\alpha z -\bar{\alpha} \overline{ z }}{ z ^2-(\overline{ z })^2-112 i }=1\right\}$ and $B =\{ z \in C :| z +3 i |=4\}$ Then $\sum_{z \in A \cap B}(\operatorname{Re} z-\operatorname{Im} z)$ is equal to $...............$.
If $z$ is a complex number such that $|z - \bar{z}| = 2$ and $|z + \bar{z}| = 4 $, then which of the following is always incorrect -
Let $z$ and $w$ be the two non-zero complex numbers such that $|z|\, = \,|w|$ and $arg\,z + arg\,w = \pi $. Then $z$ is equal to
The inequality $|z - 4|\, < \,|\,z - 2|$represents the region given by