2018-2019学年人教B版 学修2-2 3.2.2 复数的乘法和除法 教案(1)
2018-2019学年人教B版  学修2-2   3.2.2   复数的乘法和除法 教案(1)第3页

又a1a2-b1b2=a2a1-b2b1,b1a2+a1b2=b2a1+a2b1.

∴z1z2=z2z1.

(2)z1(z2+z3)=z1z2+z1z3

证明:设z1=a1+b1i,z2=a2+b2i,z3=a3+b3i(a1,a2,a3,b1,b2,b3∈R).

∵(z1z2)z3=[(a1+b1i)(a2+b2i)](a3+b3i)=[(a1a2-b1b2)+(b1b2+a1b2)i](a3+b3i)

=[(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:

z1(z2z3)=(a1a2a3-b1b2a3-b1a2b3-a1b2b3)+(b1a2a3+a1b2a3+a1a2b3-b1b2b3)i,

∴(z1z2)z3=z1(z2z3).

(3)z1(z2+z3)=z1z2+z1z3.

证明:设z1=a1+b1i,z2=a2+b2i,z3=a3+b3i(a1,a2,a3,b1,b2,b3∈R).

∵z1(z2+z3)=(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.

z1z2+z1z3=(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

∴z1(z2+z3)=z1z2+z1z3.

六、知识应用,深化理解

例1计算:(5-6i)+(-2-i)-(3+4i)

解:(5-6i)+(-2-i)-(3+4i)=(5-2-3)+(-6-1-4) i=-11 i

例2计算:(1-2i)+(-2+3i)+(3-4i)+(-4+5i)+...+(-2002+2003i)+(2003-2004i)

解法一:原式=(1-2+3-4+...-2002+2003)+(-2+3-4+5+...+2003-2004i)=(2003-1001)+(1001-2004)i=1002-1003i.

解法二:∵(1-2i)+(-2+3i)=-1+i,

(3-4i)+(-4+5i)=-1+i,

......

(2001-2002i)+(-2002+2003)i=-1+i.