If $1, \omega, \omega^2$ are the cube roots of unity, $n \in N$ and $n>2$ then the least value of $n$ such that $1+\omega$ is a root of $x^n-x=0$ is

Let $z$ be a complex number such that $|z|-z=2+i$, where $i=\sqrt{-1}$. Then, $|z|=$

If the amplitude of $z-2-3 i$ is $\pi / 4$, then the locus of $z=x+i y$ is

For $n>1$ and $n \in \mathbf{N}$, if $z_1, z_2, \ldots, z_n$ are the roots of the equation $(z+1)^n=z^n$, then $\sum_{i=1}^n \frac{\cot ^{-1}\left(2\left|\operatorname{Im} z_i\right|\right)-1}{2 \operatorname{Re} z_i}=$