Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
z = a + bi\\
z = \frac{{z' + 1}}{{z' - 1}}\\
\Leftrightarrow z\left( {z' - 1} \right) = z' + 1\\
\Leftrightarrow z.z' - z = z' + 1\\
\Leftrightarrow \left( {z - 1} \right)z' = z + 1\\
\Leftrightarrow z' = \frac{{z + 1}}{{z - 1}}\\
\Leftrightarrow z' = \frac{{a + bi + 1}}{{a + bi - 1}}\\
\Leftrightarrow z' = \frac{{\left( {a + 1} \right) + bi}}{{\left( {a - 1} \right) + bi}}\\
\Leftrightarrow z' = \frac{{\left[ {\left( {a + 1} \right) + bi} \right]\left[ {\left( {a - 1} \right) - bi} \right]}}{{\left[ {\left( {a - 1} \right) + bi} \right]\left[ {\left( {a - 1} \right) - bi} \right]}}\\
\Leftrightarrow z' = \frac{{\left( {{a^2} - 1} \right) - \left( {a + 1} \right)bi + \left( {a - 1} \right)bi - {{\left( {bi} \right)}^2}}}{{{{\left( {a - 1} \right)}^2} - {{\left( {bi} \right)}^2}}}\\
\Leftrightarrow z' = \frac{{\left( {{a^2} - 1} \right) - 2bi + {b^2}}}{{{{\left( {a - 1} \right)}^2} + {b^2}}}\\
\Leftrightarrow z' = \frac{{{a^2} + {b^2} - 1}}{{{{\left( {a - 1} \right)}^2} + {b^2}}} - \frac{{2b}}{{{{\left( {a - 1} \right)}^2} + {b^2}}}.i
\end{array}\)
