题目内容
18.若复数z满足z2=$\overline{z}$,则复数z的个数为4个.分析 设z=a+bi(a,b∈R),由于复数z满足z2=$\overline{z}$,可得a2-b2+2abi=a-bi,利用复数相等可得:$\left\{\begin{array}{l}{{a}^{2}-{b}^{2}=a}\\{2ab=-b}\end{array}\right.$,解出即可.
解答 解:设z=a+bi(a,b∈R),∵复数z满足z2=$\overline{z}$,
∴(a+bi)2=a-bi,
化为a2-b2+2abi=a-bi,
∴$\left\{\begin{array}{l}{{a}^{2}-{b}^{2}=a}\\{2ab=-b}\end{array}\right.$,
解得$\left\{\begin{array}{l}{b=0}\\{a=0或1}\end{array}\right.$,或$\left\{\begin{array}{l}{b=±\frac{\sqrt{3}}{2}}\\{a=\frac{1}{2}}\end{array}\right.$.
∴z=0,或1,或$\frac{1}{2}±\frac{\sqrt{3}}{2}i$,
故答案为:4.
点评 本题考查了复数运算法则、复数相等、方程组的解法,考查了推理能力与计算能力,属于基础题.
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