题目内容

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.

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.

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