Chọn B
Cách 1: Với \(z\ne 0\), ta có
\(z+\frac{1}{z} =\sqrt{3} \Leftrightarrow z^{2} -\sqrt{3} z+1=0 \Leftrightarrow \left[\begin{array}{l} {z=\frac{\sqrt{3} }{2} +\frac{1}{2} i} \\ {z=\frac{\sqrt{3} }{2} -\frac{1}{2} i} \end{array}\right. \)
Hay \(\left[\begin{array}{l} {z=\cos \frac{\pi }{6} +i\sin \frac{\pi }{6} } \\ {z=\cos \frac{\pi }{6} -i\sin \frac{\pi }{6} } \end{array}\right. \)
Với \(z=\cos \frac{\pi }{6} +i\sin \frac{\pi }{6} \Rightarrow z^{2016} =\cos \frac{2016\pi }{6} +i\sin \frac{2016\pi }{6} =1\)
Với \(z=\cos \frac{\pi }{6} -i\sin \frac{\pi }{6} \Rightarrow z^{2016} =\cos \frac{2016\pi }{6} -i\sin \frac{2016\pi }{6} =1\)
Vậy \(z^{2016} =1\Rightarrow T=2\)
Cách 2:
\(z+\frac{1}{z} =\sqrt{3} \Rightarrow z^{2} +\frac{1}{z^{2} } =1\)
Đặt \( t=z^{2} \Rightarrow t+\frac{1}{t} =1\Rightarrow t^{3} +\frac{1}{t^{3} } +3\left(t+\frac{1}{t} \right)=1\)
Do đó \(t^{3} +\frac{1}{t^{3} } +2=0\Rightarrow t^{6} +2t^{3} +1=0\Leftrightarrow t^{3} =-1\Rightarrow z^{6} =-1\)
\(T=z^{2016} +\frac{1}{z^{2016} } =\left(z^{6} \right)^{336} +\frac{1}{\left(z^{6} \right)^{336} } =2.\)
trannhat900 
phamngoctienpy1987844 
vxh2k9850 
Nqoc_baka 