Let $z$ be a complex number. Then the angle between vectors $z$ and $ - iz$ is

If $z $ is a complex number of unit modulus and  argument $\theta$, then ${\rm{arg}}\left( {\frac{{1 + z}}{{1 + (\bar z)}}} \right)$ equals.

If $z=x+i y, x y \neq 0$, satisfies the equation $z^2+i \bar{z}=0$, then $\left|z^2\right|$ is equal to:

If $|{z_1}|\, = \,|{z_2}|$ and $arg\,\,\left( {\frac{{{z_1}}}{{{z_2}}}} \right) = \pi $, then ${z_1} + {z_2}$ is equal to

If ${z_1}$ and ${z_2}$ are two non-zero complex numbers such that $|{z_1} + {z_2}| = |{z_1}| + |{z_2}|,$then arg $({z_1}) - $arg $({z_2})$ is equal to

If $z = x + iy\, (x, y \in R,\, x \neq \, -1/2)$ , the number of values of $z$ satisfying ${\left| z \right|^n}\, = \,{z^2}{\left| z \right|^{n - 2}}\, + \,z{\left| z \right|^{n - 2}}\, + \,1\,.\,\left( {n \in N,n > 1} \right)$ is