Let $z$ satisfy $\left| z \right| = 1$ and $z = 1 - \vec z$.
Statement $1$ : $z$ is a real number
Statement $2$ : Principal argument of $z$ is $\frac{\pi }{3}$
Statement $1$ is true Statement $2$ is true;
Statement $2$ is a correct explanation for Statement $1$.
Statement $1$ is false; Statement $2$ is true
Statement $1$ is true, Statement $2$ is false
Statement $1$ is true; Statement $2$ is true;
Statement $2$ is not a correct explanation for Statement $1$
The inequality $|z - 4|\, < \,|\,z - 2|$represents the region given by
If $\alpha$ denotes the number of solutions of $|1-i|^x=2^x$ and $\beta=\left(\frac{|z|}{\arg (z)}\right)$, where $z=\frac{\pi}{4}(1+i)^4\left(\frac{1-\sqrt{\pi i}}{\sqrt{\pi}+i}+\frac{\sqrt{\pi}-i}{1+\sqrt{\pi} \mathrm{i}}\right), i=\sqrt{-1}$, then the distance of the point $(\alpha, \beta)$ from the line $4 x-3 y=7$ is
If ${z_1}{\rm{ and }}{z_2}$ be complex numbers such that ${z_1} \ne {z_2}$ and $|{z_1}|\, = \,|{z_2}|$. If ${z_1}$ has positive real part and ${z_2}$ has negative imaginary part, then $\frac{{({z_1} + {z_2})}}{{({z_1} - {z_2})}}$may be
If complex number $z = x + iy$ is taken such that the amplitude of fraction $\frac{{z - 1}}{{z + 1}}$ is always $\frac{\pi }{4}$, then
The product of two complex numbers each of unit modulus is also a complex number, of