If $a > 0$ and $z = \frac{{{{\left( {1 + i} \right)}^2}}}{{a - i}}$, has magnitude $\sqrt {\frac{2}{5}} $, then $\bar z$ is equal to:

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,

If ${z_1},{z_2} \in C$, then $amp\,\left( {\frac{{{{\rm{z}}_{\rm{1}}}}}{{{{{\rm{\bar z}}}_{\rm{2}}}}}} \right) = $

$\left| {(1 + i)\frac{{(2 + i)}}{{(3 + i)}}} \right| = $

The conjugate of a complex number is $\frac{1}{{i - 1}}$ then that complex number is

If $z = 3 + 5i,\,\,{\rm{then }}\,{z^3} + \bar z + 198 = $