Cho so phuc z=a+bi(a,b thuoc R) thoa z×z ngaq -12/z/+(z-z ngaq)=13-10i.Tinh S=a+b

$z=a+bi,\,\overline{z}=a-bi,\,|z|=\sqrt{a^2+b^2}\\\to z.\overline{z}-12|z|+(z-\overline{z})=13-10i\\\to (a+bi)(a-bi)-12\sqrt{a^2+b^2}+(a+bi-a+bi)=13-10i\\\to a^2+b^2-12\sqrt{a^2+b^2}+2bi=13-10i\\\to \begin{cases}a^2+b^2-12\sqrt{a^2+b^2}=13\\2b=-10\end{cases}\\\text{Đặt }\sqrt{a^2+b^2}=t\,\,(t\ge0)\\\to \begin{cases}t^2-12t-13=0\\b=-5\end{cases}\to \begin{cases}\left[\begin{array}{l}t=-1\text{ (loại)}\\t=13\text{ (thỏa mãn)}\end{array}\right.\\b=-5\end{cases}\\\to \begin{cases}\sqrt{a^2+b^2}=13\\b=-5\end{cases}\to \begin{cases}a^2+b^2=169\\b=-5\end{cases}\\\to \begin{cases}a^2+25=169\\b=-5\end{cases}\to\begin{cases}a^2=144\\b=-5\end{cases}\\\to \begin{cases}a=12\\b=-5\end{cases}\to S=a+b=12-5=7$