Cho số phức $z = a + bi$ $\left( {a,\,b \in \mathbb{R},\,\,\,a > 0} \right)$ thỏa mãn $\left| {z - 1 + 2i} \right| = 5$ và $z.\bar z = 10$. Tính $P = a - b$

Theo bài ra, ta có $\left\{ \begin{array}{l}\left| {z - 1 + 2i} \right| = 5\\z.\bar z = 10\end{array} \right. \Leftrightarrow \,\,\left\{ \begin{array}{l}\left| {\left( {a - 1} \right) + \left( {b + 2} \right)i} \right| = 5\\\left( {a + bi} \right)\left( {a - bi} \right) = 10\end{array} \right. \Leftrightarrow \,\,\left\{ \begin{array}{l}{\left( {a - 1} \right)^2} + {\left( {b + 2} \right)^2} = 25\\{a^2} + {b^2} = 10\end{array} \right.$

$\begin{array}{l} \Leftrightarrow \,\,\left\{ \begin{array}{l}{a^2} + {b^2} = 10\\{a^2} + {b^2} - 2a + 4b = 20\end{array} \right. \Leftrightarrow \,\,\left\{ \begin{array}{l}2a - 4b =  - \,10\\{a^2} + {b^2} = 10\end{array} \right. \Leftrightarrow \,\,\left\{ \begin{array}{l}a = 2b - 5\\{a^2} + {b^2} = 10\end{array} \right.\\ \Rightarrow \,\,\left\{ \begin{array}{l}a = 2b - 5\\{\left( {2b - 5} \right)^2} + {b^2} = 10\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 2b - 5\\\left[ \begin{array}{l}b = 1\\b = 3\end{array} \right.\end{array} \right. \Rightarrow \left[ \begin{array}{l}\,\,\left\{ \begin{array}{l}a =  - \,3\\b = 1\end{array} \right.\left( {ktm} \right)\\\left\{ \begin{array}{l}a = 1\\b = 3\end{array} \right.\;\;\left( {tm} \right)\end{array} \right..\\ \Rightarrow P = a - b = 1 - 3 =  - 2.\end{array}$