Ta có

$\begin{align}  & {{\log }_{4}}\left( n-3 \right)+{{\log }_{2}}\sqrt{n+9}=3\left( n>3 \right) \\  & \Leftrightarrow 2.{{\log }_{4}}\left( n-3 \right)+2{{\log }_{2}}\sqrt{n+9}=6 \\ \end{align}$

$\begin{align}  & \Leftrightarrow 2.\frac{1}{2}{{\log }_{2}}\left( n-3 \right)+{{\log }_{2}}\left( n+9 \right)=6 \\  & \Leftrightarrow {{\log }_{2}}\left[ \left( n-3 \right)\left( n+9 \right) \right]=6 \\  & \Leftrightarrow \left( n-3 \right)\left( n+9 \right)=64\Leftrightarrow {{n}^{2}}+6n-27-64=0 \\ \end{align}$

$\Leftrightarrow \left[ \begin{align}  & n=7 \\  & n=-13 \\ \end{align} \right.$

Suy ra $n=7$.

Ta có

$z={{\left( \frac{2-2\sqrt{3}i}{\sqrt{3}-i} \right)}^{7}}={{\left( \sqrt{3}-i \right)}^{7}}={{\left[ 2\left( \cos \frac{-\pi }{6}+i\sin \frac{-\pi }{6} \right) \right]}^{7}}$$=128\left( \cos \frac{-7\pi }{6}+i\sin \frac{-7\pi }{6} \right)=-64\sqrt{3}+64i$

Ta chọn phương án B.