Chọn B
Gọi \(z_{1} =a+bi,\, z_{2} =c+di\, \, \left(a,b,c,d\in {\rm R}^{*} \right)\)
\(\frac{z_{1} +z_{2} }{z_{1} -z_{2} } =\frac{\left(a+bi\right)+\left(c+di\right)}{\left(a+bi\right)-\left(c+di\right)} =\frac{\left(a+c\right)+\left(b+d\right)i}{\left(a-c\right)+\left(b-d\right)i} =\frac{\left[\left(a+c\right)+\left(b+d\right)i\right]\left[\left(a-c\right)-\left(b-d\right)i\right]}{\left(a-c\right)^{2} +\left(b-d\right)^{2} } \)
\(=\frac{\left(a^{2} -c^{2} \right)+\left(b^{2} -d^{2} \right)}{\left(a-c\right)^{2} +\left(b-d\right)^{2} } +\frac{2\left(ad-bc\right)}{\left(a^{2} -c^{2} \right)+\left(b^{2} -d^{2} \right)} .i\)
Vì \(\frac{z_{1} +z_{2} }{z_{1} -z_{2} }\) là một số thuần ảo nên
\(\left(a^{2} -c^{2} \right)+\left(b^{2} -d^{2} \right)=0\Leftrightarrow a^{2} +b^{2} =c^{2} +d^{2} \Leftrightarrow \left|z_{1} \right|=\left|z_{2} \right|.\)
Từ đó suy ra \(T=\left|\frac{z_{1} }{z_{2} } \right|=\frac{\left|z_{1} \right|}{\left|z_{2} \right|} =1.\)