If $(3 + i)z = (3 - i)\bar z,$then complex number $z$ is

Lets $S=\{z \in C:|z-1|=1$ and $(\sqrt{2}-1)(z+\bar{z})-i(z-\bar{z})=2 \sqrt{2}\}$. Let $\mathrm{z}_1, \mathrm{z}_2$ $\in S$ be such that $\left|z_1\right|=\max _{z \in S}|z|$ and $\left|z_2\right|=\min _{z \in S}|z|$. Then $\left|\sqrt{2} z_1-z_2\right|^2$ equals :

The inequality $|z - 4|\, < \,|\,z - 2|$represents the region given by

If $z_{1}=2-i, z_{2}=1+i,$ find $\left|\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+1}\right|$

If for $z=\alpha+i \beta,|z+2|=z+4(1+i)$, then $\alpha+\beta$ and $\alpha \beta$ are the roots of the equation

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