Môđun của số phức z thỏa mãn |z-1|=5 và 17(z+z)-5.z.z ngang| =0 bằng:

Đáp án B

Đặt $z=a+bi\left(a;b\in ℝ\right)$ .

Ta có: $\begin{array}{l}\left\{\begin{array}{l}\left|z-1\right|=5\\ 17\left(z+\overline{z}\right)-5.z.\overline{z}=0\end{array}\right.\iff \left\{\begin{array}{l}{\left(a-1\right)}^2+b^2=25\\ 17.2a-5\left(a^2+b^2\right)=0\end{array}\right.\iff \left\{\begin{array}{l}(a^2+b^2)-2a-24=0\\ 17.2a-5\left(a^2+b^2\right)=0\end{array}\right.\\ \iff \left\{\begin{array}{l}5\left[\left(a^2+b^2\right)-2a-24\right]=0\\ 17.2a-5\left(a^2+b^2\right)=0\end{array}\right.\iff \left\{\begin{array}{l}34a+5\left(-2a-24\right)=0\\ 5\left(a^2+b^2\right)=17.2a\end{array}\right.\iff \left\{\begin{array}{l}a=5\\ a^2+b^2=34\end{array}\right.\end{array}$

Suy ra: $\left|z\right|=\sqrt{a^2+b^2}=\sqrt{34}$ .