,

\begin{eqnarray*}
5z_{1}+2z_{2} & = & 5\\
5z_{1} & = & 5-2z_{2}\\
\frac{1}{5z_{1}} & = & \frac{1}{5-2z_{2}}\\
\frac{25}{5z_{1}} & = & \frac{25}{5-2z_{2}}\\
\frac{5}{z_{1}} & = & \frac{25}{5-2z_{2}}\\
5z_{1}^{-1} & = & \frac{25}{5-2z_{2}}
\end{eqnarray*}

,

take คอนจูเกททั้งสองข้างของ $\bar{z}_2 = 1+2i$ จะได้ 

\begin{eqnarray*}
\bar{z}_{2} & = & 1+2i\\
\overline{\left(\bar{z}_{2}\right)} & = & \overline{1+2i}\\
\bar{\bar{z}}_{2} & = & 1-2i\\
z_{2} & = & 1-2i
\end{eqnarray*}

แทนค่า $z_2 = 1 - 2i$ ลงในสมการ $\left| 5z_1^{-1} \right|$ แล้วคำนวณค่า $\left| 5 z_1^{-1} \right|$

\begin{eqnarray*}
\left|5z_{1}^{-1}\right| & = & \left|\frac{25}{5-2z_{2}}\right|\\
 & = & \left|\frac{25}{5-2\left(1-2i\right)}\right|\\
 & = & \frac{\left|25\right|}{\left|5-2+4i\right|}\\
 & = & \frac{25}{\left|3+4i\right|}
\end{eqnarray*}

คำนวณ $\left| 3 + 4i \right|$

\begin{eqnarray*}
\left|5z_{1}^{-1}\right| & = & \frac{25}{\left|3+4i\right|}\\
 & = & \frac{25}{\sqrt{3^{2}+4^{2}}}\\
 & = & \frac{25}{\sqrt{25}}\\
 & = & \frac{25}{5}\\
 & = & 5
\end{eqnarray*}