Conjugate of $1 + i$ is

If $|z|\, = 1,(z \ne - 1)$and $z = x + iy,$then $\left( {\frac{{z - 1}}{{z + 1}}} \right)$ is

Let $S=\left\{Z \in C: \bar{z}=i\left(z^2+\operatorname{Re}(\bar{z})\right)\right\}$. Then $\sum_{z \in S}|z|^2$ is equal to

Let $z$ =${i^{2i}}$ , then $|z|$ is (where $i$ =$\sqrt { - 1}$ )

If the set $\left\{\operatorname{Re}\left(\frac{z-\bar{z}+z \bar{z}}{2-3 z+5 \bar{z}}\right): z \in C , \operatorname{Re}(z)=3\right\}$ is equal to the interval $(\alpha, \beta]$, then $24(\beta-\alpha)$ 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,