Cho số phức \(z=a+bi\left( a,b\in \mathbb{R} \right)\) thỏa mãn phương trình \(\frac{\left( \left| z \right|-1 \right)\left( 1+iz \right)}{z-\frac{1}{z}}=i.\) Tính P=a+b.

* Hướng dẫn giải

\(\begin{array}{l} \frac{{\left( {\left| z \right| - 1} \right)\left( {1 + iz} \right)}}{{z - \frac{1}{{\overline z }}}} = i \Leftrightarrow \frac{{\left( {\left| z \right| - 1} \right)\left( {1 + iz} \right)\overline z }}{{z\overline z - 1}} = i{\rm{ }}\left( {\left| z \right| \ne 1} \right)\\ \Leftrightarrow \frac{{\left( {\left| z \right| - 1} \right)\left( {1 + iz} \right)\overline z }}{{{{\left| z \right|}^2} - 1}} = i \Leftrightarrow \frac{{\left( {1 + iz} \right)\overline z }}{{\left| z \right| + 1}} = i\\ \Leftrightarrow \overline z + i{\left| z \right|^2} = i\left( {\left| z \right| + 1} \right) \Leftrightarrow a - bi + \left( {{a^2} + {b^2}} \right)i = i\left( {\sqrt {{a^2} + {b^2}} + 1} \right)\\ \Leftrightarrow a + \left( { - b + {a^2} + {b^2}} \right)i = i\left( {\sqrt {{a^2} + {b^2}} + 1} \right) \Leftrightarrow \left\{ \begin{array}{l} a = 0\\ {b^2} - b = \left| b \right| + 1 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} a = 0\\ \left[ \begin{array}{l} \left\{ \begin{array}{l} b < 0\\ b = \pm 1\left( L \right) \end{array} \right.\\ \left\{ \begin{array}{l} b > 0\\ {b^2} - 2b - 1 = 0 \end{array} \right. \end{array} \right. \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} a = 0\\ \left[ \begin{array}{l} b = 1 + \sqrt 2 \left( N \right)\\ b = 1 - \sqrt 2 \left( L \right) \end{array} \right. \end{array} \right.\\ \Rightarrow P = a + b = 1 + \sqrt 2 . \end{array}\)