Lời giải:Gọi z=a+bi ; \(a,b\in \mathbb{R};\,{{i}^{2}}=-1\); a là số nguyên. Theo đề ta có
\(|z|-2\overline{z}=-7+3i+z\)
\(\Leftrightarrow \sqrt{{{a}^{2}}+{{b}^{2}}}-2a+2bi=-7+3i+a+bi\)
\(\Leftrightarrow (\sqrt{{{a}^{2}}+{{b}^{2}}}-2a)+2bi=(-7+a)+(3+b)i\)
\( \Leftrightarrow \left\{ \begin{array}{l}
\sqrt {{a^2} + {b^2}} - 2a = - 7 + a\\
2b = 3 + b
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\sqrt {{a^2} + 9} = 3a - 7\\
b = 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left\{ \begin{array}{l}
a \ge \frac{7}{3}\\
8{a^2} - 42a + 40 = 0
\end{array} \right.\\
b = 3
\end{array} \right.\)
\( \Leftrightarrow \left\{ \begin{array}{l}
a \ge \frac{7}{3}\\
\left[ \begin{array}{l}
a = 4\\
a = \frac{5}{4}
\end{array} \right.\\
b = 3
\end{array} \right.\) \(\Leftrightarrow \left\{ \begin{array}{l}
a = 4\\
b = 3
\end{array} \right.\)
Khi đó z = 4 + 3i
Vậy \(w = 1 - z + {z^2} = 4 + 21i \Rightarrow \left| w \right| = \sqrt {457} \).
