Đáp án:
$\begin{array}{l}
z = a + bi\\
a)\left| {2 + z} \right| < \left| {2 - z} \right|\\
\Rightarrow \sqrt {{{\left( {2 + a} \right)}^2} + {b^2}} < \sqrt {{{\left( {2 - a} \right)}^2} + {b^2}} \\
\Rightarrow {a^2} + 4a + 4 + {b^2} < {a^2} - 4a + 4 + {b^2}\\
\Rightarrow a < 0\\
b)\left| {z + \overline z + 3} \right| = 4\\
\Rightarrow \left| {a + bi + a - bi + 3} \right| = 4\\
\Rightarrow \left| {2a + 3} \right| = 4\\
\Rightarrow \left[ \begin{array}{l}
2a + 3 = 4\\
2a + 3 = - 4
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
a = \frac{1}{2}\\
a = - \frac{7}{2}
\end{array} \right.
\end{array}$
