Chọn B
Gọi \(z_{1} =a_{1} +b_{1} i;z_{2} =a_{2} +b_{2} i.\)
Ta có:
\(\left|z_{2} \right|=\left|z_{1} -z_{2} \right|\)\(\Leftrightarrow a_{2}^{2} +b_{2}^{2} =(a_{1} -a_{2} )^{2} +(b_{1} -b_{2} )^{2} \)
\(\Leftrightarrow a_{2}^{2} +b_{2}^{2} =a_{1}^{2} +a_{2}^{2} +b_{1}^{2} +b_{2}^{2} -2(a_{1} a_{1} +b_{1} b_{2} )\)
\(\Leftrightarrow a_{1}^{2} +b_{1}^{2} =2(a_{1} a_{1} +b_{1} b_{2} )=\left|z_{1} \right|^{2}\)
\(\begin{array}{l} {\frac{z_{1} }{z_{2} } +\frac{z_{2} }{z_{1} } =\frac{z_{1} .\overline{z_{2} }}{\left|z_{2} \right|^{2} } +\frac{z_{2} .\overline{z_{1} }}{\left|z_{1} \right|^{2} } =\frac{z_{1} .\overline{z_{2} }+z_{2} .\overline{z_{1} }}{\left|z_{1} \right|^{2} } =\frac{(a_{1} +b_{1} i)(a_{2} -b_{2} i)+(a_{1} -b_{1} i)(a_{2} +b_{2} i)}{\left|z_{1} \right|^{2} } } \\ {{\rm \; \; \; \; \; \; \; \; \; \; \; \; \; =}\frac{a_{1} a_{2} -a_{1} b_{2} i+b_{1} a_{2} i+b_{1} b_{2} +a_{1} a_{2} +a_{1} b_{2} i-b_{1} a_{2} i+b_{1} b_{2} }{\left|z_{1} \right|^{2} } =\frac{2(a_{1} a_{2} +b_{1} b_{2} )}{\left|z_{1} \right|^{2} } =1} \end{array}\)
Vậy: \(A=\left(\frac{z_{1} }{z_{2} } \right)^{4} +\left(\frac{z_{2} }{z_{1} } \right)^{4} =\left(\left(\frac{z_{1} }{z_{2} } \right)^{2} +\left(\frac{z_{2} }{z_{1} } \right)^{2} \right)^{2} -2\)
\(=\left(\left(\frac{z_{1} }{z_{2} } +\frac{z_{2} }{z_{1} } \right)^{2} -2\right)^{2} -2=-1.\)