已知復數(shù)z=(a2+a)+(a+2)i(a∈R).
(1)若復數(shù)z為實數(shù),求實數(shù)a的值;
(2)若復數(shù)z的共軛復數(shù)對應(yīng)的點在第四象限,求實數(shù)a的取值范圍.
解:(1)復數(shù)z=(a
2+a)+(a+2)i(a∈R)
若復數(shù)z為實數(shù),則a+2=0,所以a=-2;
(2)復數(shù)z的共軛復數(shù)是

=(a
2+a)-(a+2)i(a∈R)
若復數(shù)z的共軛復數(shù)對應(yīng)的點在第四象限,
則

,解①得:a<-1或a>0,解②得:a>-2,
所以不等式組的解集為{a|-2<a<-1或a>0}.
所以復數(shù)z的共軛復數(shù)對應(yīng)的點在第四象限的實數(shù)a的取值范圍是{a|-2<a<-1或a>0}.
分析:(1)復數(shù)z=(a
2+a)+(a+2)i(a∈R)是實數(shù),只要其虛部為0即可;
(2)寫出復數(shù)z的共軛復數(shù),對應(yīng)的點在第四象限,說明其實部大于0,虛部小于0,列不等式求解a的取值范圍.
點評:本題考查了復數(shù)的基本概念,關(guān)鍵是讀懂題意,把問題轉(zhuǎn)化為方程或不等式組求解,此題是基礎(chǔ)題.