The conjugate of $\frac{{{{(2 + i)}^2}}}{{3 + i}},$ in the form of $a + ib$, is
$\frac{{13}}{2} + i\,\left( {\frac{{15}}{2}} \right)$
$\frac{{13}}{{10}} + i\left( {\frac{{ - 15}}{2}} \right)$
$\frac{{13}}{{10}} + i\,\left( {\frac{{ - 9}}{{10}}} \right)$
$\frac{{13}}{{10}} + i\,\left( {\frac{9}{{10}}} \right)$
Argument of $ - 1 - i\sqrt 3 $ is
A real value of $x$ will satisfy the equation $\left( {\frac{{3 - 4ix}}{{3 + 4ix}}} \right) = $ $\alpha - i\beta \,(\alpha ,\beta \,{\rm{real),}}$ if
If $|z_1|=1, \, |z_2| =2, \,|z_3|=3$ and $|9z_1z_2 + 4z_1z_3+z_2z_3| =12$ then the value of $|z_1+z_2+z_3|$ is equal to :-
If complex numbers $(x -2y) + i(3x -y)$ and $(2x -y) + i(x -y + 6)$ are conjugates of each other, then $|x + iy|$ is $(x,y \in R)$
If $|z|\, = 1$ and $\omega = \frac{{z - 1}}{{z + 1}}$ (where $z \ne - 1)$, then ${\mathop{\rm Re}\nolimits} (\omega )$ is