If  $z_1 = a + ib$ and $z_2 = c + id$ are complex numbers such that   $| z_1 | = | z_2 |=1$ and  $R({z_1}\overline {{z_2}} ) = 0$, then the pair of complex numbers $w_1 = a + ic$ and $w_2 = b + id$ satisfies

If $z=\frac{1}{2}-2 i$, is such that $|z+1|=\alpha z+\beta(1+i), i=\sqrt{-1}$ and $\alpha, \beta \in R \quad$, then $\alpha+\beta$ is equal to

The conjugate of complex number $\frac{{2 - 3i}}{{4 - i}},$ is

The complex numbers $sin\ x + i\ cos\ 2x$ and $cos\ x\ -\ i\ sin\ 2x$ are conjugate to each other, for

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

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,