Gọi số phức \(z = a + bi,\,\,a,b \in R\)
\(\begin{array}{l}
\left\{ \begin{array}{l}
|z - 2i| = |z|\\
|z - i| = |z - 1|
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
|a + bi - 2i| = |a + bi|\\
|a + bi - i| = |a + bi - 1|
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{a^2} + {(b - 2)^2} = {a^2} + {b^2}\\
{a^2} + {(b - 1)^2} = {(a - 1)^2} + {b^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
- 4b + 4 = 0\\
- 2b = - 2a
\end{array} \right.\\
\Leftrightarrow a = b = 1
\end{array}\)
Vậy \(z = 1 + i\)
-- Mod Toán 12