Cho số phức \(z=a+bi\left( a,\,b\in \mathbb{R},\,a>0 \right)\) thỏa \(z.\bar{z}-12\left| z \right|+\left( z-\bar{z} \right)=13-10i\). Tính S=a+b.

* Hướng dẫn giải

\(z.\bar z - 12\left| z \right| + \left( {z - \bar z} \right) = 13 - 10i\)

\( \Leftrightarrow {a^2} + {b^2} - 12\sqrt {{a^2} + {b^2}}  + 2bi = 13 - 10i\)

\( \Leftrightarrow \left\{ \begin{array}{l} {a^2} + {b^2} - 12\sqrt {{a^2} + {b^2}} = 13\\ 2b = - 10 \end{array} \right.\)

\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l} {a^2} + 25 - 12\sqrt {{a^2} + 25} = 13\\ b = - 5 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} \left[ \begin{array}{l} \sqrt {{a^2} + 25} = 13\\ \sqrt {{a^2} + 25} = - 1\left( {VN} \right) \end{array} \right.\\ b = - 5 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} a = \pm 12\\ b = - 5 \end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l} a = 12\\ b = - 5 \end{array} \right. \end{array}\)

Vậy S = a + b = 7