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

If $a_k = \cos \alpha_k + i \sin \alpha_k$ for $k = 1, 2, 3$ and $a_1, a_2, a_3$ are the roots of the equation $x^3 + bx + c = 0$,then the real part of $b$ is:

If $2 + 3i$ is one of the roots of the equation $2x^3 - 9x^2 + kx - 13 = 0,$ where $k \in R,$ then the real root of this equation:

If $\sqrt{-3-4 i}=re^{i \theta}$,then $r^2 \tan \theta=$

$\frac{{{{( - 1 + i\sqrt 3 )}^{15}}}}{{{{(1 - i)}^{20}}}} + \frac{{{{( - 1 - i\sqrt 3 )}^{15}}}}{{{{(1 + i)}^{20}}}} = \dots$

If a complex number $z$ is such that $(7+i)(z+\bar{z})-(4+i)(z-\bar{z})+116i=0$,then $z\bar{z}=$