The solution of the equation $|z| - z = 1 + 2i$ is

Let $a = lm\left( {\frac{{1 + {z^2}}}{{2iz}}} \right)$, where $z$ is any non-zero complex number. The set $A = \{ a:\left| z \right| = 1\,and\,z \ne  \pm 1\} $ is equal to

Let $a \neq b$ be two non-zero real numbers.Then the number of elements in the set $X =\left\{ z \in C : \operatorname{Re}\left(a z^2+ bz \right)= a \text { and }\operatorname{Re}\left(b z^2+ az \right)= b \right\}$ is equal to

If ${z_1}$ and ${z_2}$ are two complex numbers, then $|{z_1} - {z_2}|$ is

If $z=\frac{1}{2}-2 i$, is such that $|z+1|=\alpha z+\beta(1+i), i=\sqrt{-1}$ and $\alpha, \beta \in R \quad$, then $\alpha+\beta$ is equal to

The amplitude of the complex number $z = \sin \alpha + i(1 - \cos \alpha )$ is