Chọn B
Ta có
\(z=\left(\frac{1+i}{\sqrt{3} +i} \right)^{6} =\frac{\left(1+i\right)^{6} }{\left(\sqrt{3} +i\right)^{6} } =\frac{\left[\left(1+i\right)^{2} \right]^{3} }{\left[\left(\sqrt{3} +i\right)^{3} \right]^{2} } =\frac{\left(2i\right)^{3} }{\left(8i\right)^{2} } =\frac{-8i}{-64} =\frac{1}{8} i.\)