Question

12．已知复数z的共轭复数为$\overline z$，且z•$\overline z-3iz=\frac{10}{1-3i}$，求复数z．

Answer 1

分析 设z=a+bi（a，b∈R），则$\overline{z}$=a-bi，代入化简$\left.\begin{array}{l}{z•\overline{z}-3iz}\end{array}\right.$，再代入z•$\overline z-3iz=\frac{10}{1-3i}$化简，利用复数相等的条件列出方程组，求出a、b的值即可求出复数z．

解答 解：设z=a+bi（a，b∈R），则$\overline{z}$=a-bi，
∴$\left.\begin{array}{l}{z•\overline{z}-3iz={a}^{2}+{b}^{2}+3b-3ai}\end{array}\right.$，
∵z•$\overline z-3iz=\frac{10}{1-3i}$，
∴$\left.\begin{array}{l}{{a}^{2}+{b}^{2}+3b-3ai=\frac{10（1+3i）}{（1-3i）（1+3i）}=1+3i}\end{array}\right.$，
则$\left.\begin{array}{l}{\left\{\begin{array}{l}{{a}^{2}+{b}^{2}+3b=1}\\{-3a=3}\end{array}\right.}\end{array}\right.$，解得$\left.\begin{array}{l}{\left\{\begin{array}{l}{a=-1}\\{b=0}\end{array}\right.}\end{array}\right.$或$\left.\begin{array}{l}{\left\{\begin{array}{l}{a=-1}\\{b=-3}\end{array}\right.}\end{array}\right.$
∴z=-1或$\left.\begin{array}{l}{z=-1-3i}\end{array}\right.$．

点评 本题考查复数代数形式的乘除运算，共轭复数和复数相等的定义，以及化简、计算能力．