2024-2025學(xué)年新教材高中數(shù)學(xué)第7章復(fù)數(shù)7.2.1復(fù)數(shù)的加減運(yùn)算及其幾何意義鞏固練習(xí)含解析新人教A版必修第二冊(cè)

7.2.1復(fù)數(shù)的加、減運(yùn)算及其幾何意義課后訓(xùn)練鞏固提升1.若復(fù)數(shù)z滿意z+i-3=3-i,則z等于()A.0 B.2iC.6 D.6-2i解析:∵z+i-3=3-i,∴z=(3-i)-(i-3)=(3+3)+(-i-i)=6-2i,故選D.答案:D2.若復(fù)數(shù)z滿意z+(3-4i)=1,則z的虛部是()A.-2 B.4 C.3 D.-4解析:z=1-(3-4i)=-2+4i,故選B.答案:B3.設(shè)m∈R,復(fù)數(shù)z=(2m2+3i)+(m-m2i)+(-1+2mi),若z為純虛數(shù),則m等于()A.-1 B.3C.12 D.-1或解析:z=(2m2+m-1)+(3+2m-m2)i.由題意得2m2+m答案:C4.在復(fù)平面內(nèi),O是原點(diǎn),OA,OC,AB表示的復(fù)數(shù)分別為-2+i,3+2i,1+5i,則BCA.2+8i B.-6-6iC.4-4i D.-4+2i解析:BC=OC-OB=(3,2)-(1,5)-(-2,1)=(4,-4).答案:C5.若|z-1|=|z+1|,則復(fù)數(shù)z對(duì)應(yīng)的點(diǎn)在()A.實(shí)軸上 B.虛軸上C.第一象限 D.其次象限解析:∵|z-1|=|z+1|,∴點(diǎn)Z到(1,0)和(-1,0)的距離相等,即點(diǎn)Z在以(1,0)和(-1,0)為端點(diǎn)的線段的中垂線上.答案:B6.已知x∈R,y∈R,(xi+x)+(yi+4)=(y-i)-(1-3xi),則x=,y=.

解析:∵x+4+(x+y)i=(y-1)+(3x-1)i,∴x+4=y答案:6117.已知z1=32a+(a+1)i,z2=-33b+(b+2)i(a,b∈R),若z1-z2=43,則a+b=.解析:∵z1-z2=32a+(a+1)i-[-33b+(b+2)i]=32a+33b+(由復(fù)數(shù)相等的條件知3解得a=2,b=1答案:38.設(shè)復(fù)數(shù)z滿意z+|z|=2+i,則z=.

解析:設(shè)z=x+yi(x,y∈R),則|z|=x2∴x+yi+x2+y2∴x+x2+y2=2,答案:34+9.計(jì)算下列各式.(1)(3-2i)-(10-5i)+(2+17i);(2)(1-2i)-(2-3i)+(3-4i)-(4-5i)+…+(2019-2020i).解:(1)原式=(3-10+2)+(-2+5+17)i=-5+20i.(2)原式=(1-2+3-4+…+2024-2025+2024)+(-2+3-4+5-…-2025+2024-2025)i=1010-1011i.10.在復(fù)平面內(nèi),A,B,C三點(diǎn)對(duì)應(yīng)的復(fù)數(shù)分別為1,2+i,-1+2i.(1)求向量AB,AC(2)推斷△ABC的形態(tài);(3)求△ABC的面積.解:(1)AB對(duì)應(yīng)的復(fù)數(shù)為2+i-1=1+i,BC對(duì)應(yīng)的復(fù)數(shù)為-1+2i-(2+i)=-3+i,AC對(duì)應(yīng)的復(fù)數(shù)為-1+2i-1=-2+2i.(2)∵|AB|=2,|B