Cho hai số phức \(z_1, z_2\) thỏa mãn \(\left| {{z_1}} \right| = 3,\left| {{z_2}} \right| = 4,\left| {{z_1} - {z_2}} \right| = \sqrt {41}

* Hướng dẫn giải

Ta có: \(\left| {{z_1}} \right| = 3,\left| {{z_2}} \right| = 4,\left| {{z_1} - {z_2}} \right| = \sqrt {41}  \Rightarrow \left\{ \begin{array}{l}
\frac{{\left| {{z_1}} \right|}}{{\left| {{z_2}} \right|}} = \frac{3}{4}\\
\frac{{\left| {{z_1} - {z_2}} \right|}}{{\left| {{z_2}} \right|}} = \frac{{\sqrt {41} }}{4}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\frac{{\left| {{z_1}} \right|}}{{\left| {{z_2}} \right|}} = \frac{3}{4}\\
\left| {\frac{{{z_1}}}{{{z_2}}} - 1} \right| = \frac{{\sqrt {41} }}{4}
\end{array} \right.\) 

\(z = \frac{{{z_1}}}{{{z_2}}} = a + bi,\left( {a,b \in R} \right) \Rightarrow \left\{ \begin{array}{l}
{a^2} + {b^2} = {\left( {\frac{3}{4}} \right)^2}\\
{\left( {a - 1} \right)^2} + {b^2} = {\left( {\frac{{\sqrt {41} }}{4}} \right)^2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{a^2} + {b^2} = \frac{9}{{16}}\\
{\left( {a - 1} \right)^2} + {b^2} = \frac{{41}}{{16}}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{b^2} = \frac{9}{{16}} - {a^2}\\
{\left( {a - 1} \right)^2} + \frac{9}{{16}} - {a^2} = \frac{{41}}{{16}}
\end{array} \right.\) 

\( \Leftrightarrow \left\{ \begin{array}{l}
{b^2} = \frac{5}{{16}}\\
a =  - \frac{1}{2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left| b \right| = \frac{{\sqrt 5 }}{4}\\
a =  - \frac{1}{2}
\end{array} \right.\) 

Vậy \(\left| b \right| = \frac{{\sqrt 5 }}{4}\)