If $\frac{{2{z_1}}}{{3{z_2}}}$ is a purely imaginary number, then $\left| {\frac{{{z_1} - {z_2}}}{{{z_1} + {z_2}}}} \right|$ =

Let $z$ be a complex number (not lying on $X$-axis) of maximum modulus such that $\left| {z + \frac{1}{z}} \right| = 1$. Then

Consider the following two statements :

Statement $I$ : For any two non-zero complex numbers $\mathrm{z}_1, \mathrm{z}_2$

$\left(\left|z_1\right|+\left|z_2\right|\right)\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq 2\left(\left|z_1\right|+\left|z_2\right|\right)$ and

Statement $II$ : If $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are three distinct complex numbers and a, b, c are three positive real numbers such that $\frac{a}{|y-z|}=\frac{b}{|z-x|}=\frac{c}{|x-y|}$, then

$\frac{\mathrm{a}^2}{\mathrm{y}-\mathrm{z}}+\frac{\mathrm{b}^2}{\mathrm{z}-\mathrm{x}}+\frac{\mathrm{c}^2}{\mathrm{x}-\mathrm{y}}=1$

Between the above two statements,

Modulus of $\left( {\frac{{3 + 2i}}{{3 - 2i}}} \right)$ is

If$z = \frac{{1 - i\sqrt 3 }}{{1 + i\sqrt 3 }},$then $arg(z) = $ ............. $^\circ$

For $a \in C$, let $A =\{z \in C: \operatorname{Re}( a +\overline{ z }) > \operatorname{Im}(\bar{a}+z)\}$ and $B=\{z \in C: \operatorname{Re}(a+\bar{z}) < \operatorname{Im}(\bar{a}+z)\}$. Then among the two statements :

$(S 1)$ : If $\operatorname{Re}(A), \operatorname{Im}(A) > 0$, then the set $A$ contains all the real numbers

$(S2)$: If $\operatorname{Re}(A), \operatorname{Im}(A) < 0$, then the set $B$ contains all the real numbers,