Đặt: \(z = a + bi,\,\,(a,b \in \mathbb{R}),\) ta có: \(a + \sqrt {{a^2} + {b^2}} + bi = 3 + 4i \Rightarrow \left\{ \begin{array}{l} b = 4\\ a + \sqrt {{a^2} + {b^2}} = 3 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} b = 4\\ a = - \frac{7}{6} \end{array} \right..\)