=[(a1a2-b1b2)a3-(b1a2+a1b2)b3]+[(b1a2+a1b2)a3+(a1a2-b1b2)b3]i
=(a1a2a3-b1b2a3-b1a2b3-a1b2b3)+(b1a2a3+a1b2b3+a1a2b3-b1b2b3)i,
同理可证:
1( 2 3)=(a1a2a3-b1b2a3-b1a2b3-a1b2b3)+(b1a2a3+a1b2a3+a1a2b3-b1b2b3)i,
∴( 1 2) 3= 1( 2 3).
(3) 1( 2+ 3)= 1 2+ 1 3.
证明:设 1=a1+b1i, 2=a2+b2i, 3=a3+b3i(a1,a2,a3,b1,b2,b3∈R).
∵ 1( 2+ 3)=(a1+b1i)[(a2+b2i)+(a3+b3i)]=(a1+b1i)[(a2+a3)+(b2+b3)i]
=[a1(a2+a3)-b1(b2+b3)]+[b1(a2+a3)+a1(b2+b3)]i
=(a1a2+a1a3-b1b2-b1b3)+(b1a2+b1a3+a1b2+a1b3)i.
1 2+ 1 3=(a1+b1i)(a2+b2i)+(a1+b1i)(a3+b3i)
=(a1a2-b1b2)+(b1a2+a1b2)i+(a1a3-b1b3)+(b1a3+a1b3)i
=(a1a2-b1b2+a1a3-b1b3)+(b1a2+a1b2+b1a3+a1b3)i
=(a1a2+a1a3-b1b2-b1b3)+(b1a2+b1a3+a1b2+a1b3)i
∴ 1( 2+ 3)= 1 2+ 1 3.
例1计算(1-2i)(3+4i)(-2+i)
【解析】(1-2i)(3+4i)(-2+i)=(11-2i) (-2+i)= -20+15i.
例2计算:
(1)(3+4i) (3-4i) ; (2)(1+ i)2.
【解析】(1)(3+4i) (3-4i) =32-(4i)2=9-(-16)=25;
(2) (1+ i)2=1+2 i+i2=1+2 i-1=2 i.
3.共轭复数:当两个复数的实部相等,虚部互为相反数时,这两个复数叫做互为共轭复数虚部不等于0的两个共轭复数也叫做共轭虚数
通常记复数的共轭复数为。
4. 复数除法定义:满足(c+di)(x+yi)=(a+bi)的复数x+yi(x,y∈R)叫复数a+bi除以复数c+di的商,记为:(a+bi)(c+di)或者
5.除法运算规则:
①设复数a+bi(a,b∈R),除以c+di(c,d∈R),其商为x+yi(x,y∈R),
即(a+bi)÷(c+di)=x+yi