Ta chọn câu C
Gọi \(z=a+bi\, \, \left(a,b\in {\rm R}\right) \Rightarrow \overline{z}=a-bi.\)
Ta có: \(\left(z+1\right)^{2} +\, \left|z-1\right|^{2} -10i=\overline{z}\, +\, 3\)
\(\Leftrightarrow \left(a+1+bi\right)^{2} +\left|a-1+bi\right|^{2} -10i=a-bi+3\)
\(\Leftrightarrow \left(a+1\right)^{2} -b^{2} +2\left(a+1\right)bi+\left(a-1\right)^{2} +b^{2} -10i=a+3-bi
\)
\(\Leftrightarrow 2a^{2} +2+2\left(ab+b-5\right)i=a+3-bi \Leftrightarrow \left\{\begin{array}{l} {2a^{2} +2=a+3} \\ {2ab+2b-10=-b} \end{array}\right. \)
\(\Leftrightarrow \left\{\begin{array}{l} {2a^{2} -a-1=0\, \, \, } \\ {2ab+3b=10\, \, \, \, \, } \end{array}\right. \Leftrightarrow \left\{\begin{array}{l} {\left[\begin{array}{l} {a=1} \\ {a=-\frac{1}{2} } \end{array}\right. \, \, \, } \\ {2ab+3b=10\, \, \, \, \, } \end{array}\right. \Leftrightarrow \left[\begin{array}{l} {\left\{\begin{array}{l} {a=1} \\ {b=2} \end{array}\right. } \\ {\left\{\begin{array}{l} {a=-\frac{1}{2} } \\ {b=5} \end{array}\right. } \end{array}\right. \)
Vậy có 2 số phức z thỏa mãn yêu cầu là \(z=1+2i\, ;\, z=-\frac{1}{2} +5i.\)