高中數(shù)學(xué)第3章數(shù)系的擴(kuò)充與復(fù)數(shù)的引入3.2復(fù)數(shù)的四則運(yùn)算自我

1、高中數(shù)學(xué) 第3章 數(shù)系的擴(kuò)充與復(fù)數(shù)的引入 3.2 復(fù)數(shù)的四則運(yùn)算自我小測(cè) 蘇教版選修2-21.若復(fù)數(shù)z 滿(mǎn)足z +i -3=3-i ,則z =_.2.(2012天津高考改編i 是虛數(shù)單位,復(fù)數(shù)7i 3i-=+_. 3.設(shè)a R ,且1i 1i 2a -+是實(shí)數(shù),則a =_. 4.已知i 為虛數(shù)單位,則 2 0121i 1i -= +_. 5.已知z 1=1+2i ,z 2=m +(m -1i ,且兩復(fù)數(shù)的乘積z 1z 2的實(shí)部和虛部為相等的正數(shù),則實(shí)數(shù)m 的值為_(kāi).6.已知復(fù)數(shù)22i (1iz +=+,則z 的共軛復(fù)數(shù)的虛部為_(kāi). 7.若復(fù)數(shù)z 滿(mǎn)足z -2i =1+z i(i 為虛數(shù)單位,則

2、z =_.8.設(shè)z 1=x +2i ,z 2=3-y i(x ,y R ,且z 1+z 2=5-6i ,則z 1-z 2=_.9.計(jì)算: (1112i 2i 22-+- ;(2(3+2i+2i ; (3(1+2i+(i +i 2+(3+4i;(4(6-3i+(3+2i-(3-4i-(-2+i. 10.已知復(fù)數(shù)2(1i3(1i2iz -+=+. (1求復(fù)數(shù)z ;(2若z 2+az +b =1-i ,求實(shí)數(shù)a ,b 的值.參考答案1答案:6-2i 解析:由已知z =(3-i+(3-i=6-2i.2答案:2-i 解析:227i (7i(3i217i 3i i 2010i 2i 3i (3i(3i9i

3、 10-+-=-+-. 3答案:-1 解析1i (1i1i (1(1i 1i 2222a a a a -+-+=+=+. 則當(dāng)此復(fù)數(shù)為實(shí)數(shù)時(shí),有a +1=0,a =-1.4答案:1 解析:21i (1ii 1i 2-=-+. 2 0121i 1i - +=(-i2 012=i 2 012=i 4×503=i 0=1. 5答案:34解析:z 1z 2=(1+2im +(m -1i =m +(m -1i +2m i -2(m -1=(2-m +(3m -1i ,由已知得2-m =3m -1>0,解得m =34. 6答案:1 解析:22i 2i (2i(i12i 1i (1i2i 2

4、22z +-=-+,1i 2z =+,即z 的共軛復(fù)數(shù)的虛部為1.7答案:13i 22-+ 解析:由已知得z -z i =1+2i , 12i (12i(1i13i 1i (1i(1i2z +-+=-+ =13i 22-+. 8答案:-1+10i 解析:z 1+z 2=(x +2i+(3-y i=(x +3+(2-y i , 又z 1+z 2=5-6i ,3526x y +=-=-,28.x y =,z1=2+2i,z2=3-8i.z1-z2=(2+2i-(3-8i=(2-3+(2+8i=-1+10i.9答案:解:(1原式=1122i22+-+ =55i 22 -;(2原式=3+(2 2i=3 ;(3原式=(1+2i+(i-1+(3+4i=(1-1+3+(2+1+4i=3+7i;(4原式=6+3-3-(-2+-3+2-(-4-1i=8+2i.10答案:解:(12i33i3i(3i(2i1i 2i2i5z-+=+-.(2把