§10.2

復(fù)數(shù)的運算【復(fù)習(xí)目標】1.理解復(fù)數(shù)代數(shù)形式的加法、減法和乘法運算.2.了解復(fù)數(shù)加法和減法運算的幾何意義.【知識回顧】1.復(fù)數(shù)代數(shù)形式的四則運算法則設(shè)z1=a+bi,z2=c+di,(a,b,c,d∈R)是任意兩個復(fù)數(shù).(1)z1+z2=(a+bi)+(c+di)=(a+c)+(b+d)i;(2)z1-z2=(a+bi)-(c+di)=(a-c)+(b-d)i;(3)z1z2=(a+bi)(c+di)=(ac-bd)+(ad+bc)i.2.復(fù)數(shù)的運算律復(fù)數(shù)的加法和乘法都滿足實數(shù)加法和乘法的所有運算律.對于任意復(fù)數(shù)z1,z2,z3,均滿足以下運算律:(1)交換律:z1+z2=z2+z1;(2)結(jié)合律:(z1+z2)+z3=z1+(z2+z3);(3)分配律:z1(z2+z3)=z1z2+z1z3.

【例題精解】【例1】已知z1=3i,z2=1-i,z3=-2+5i,計算z1-z2,z1+z2-z3.【解】

z1-z2=3i-(1-i)=(0-1)+[3-(-1)]i=-1+4i;z1+z2-z3=3i+(1-i)-(-2+5i)=[0+1-(-2)]+[3+(-1)-5]i=3-3i.【點評】根據(jù)復(fù)數(shù)的運算法則進行解答.【對點練習(xí)1】

(1+i)-2i-(2-i)=

.

【答案】-1【解析】(1+i)-2i-(2-i)=(1-0-2)+(1-2+1)i=-1.【例2】

計算:(1)(2+3i)(2-i); (2)(1+i)2.【解】(1)(2+3i)(2-i)=4-2i+6i-3i2=7+4i;(2)(1+i)2=12+2i+i2=2i.【點評】根據(jù)復(fù)數(shù)的運算法則進行解答.【對點練習(xí)2】

(1)(3+2i)(4-3i)=

;

(2)(1-2i)i=

.

【解】(1)18-i

(3+2i)(4-3i)=12-9i+8i-6i2=18-i.

(2)2+i

(1-2i)i=i-2i2=2+i.【例3】已知i為虛數(shù)單位,復(fù)數(shù)z滿足zi=-1+i,則z= (

) A.1+i B.1-i C.-1+i D.-1-i

【點評】用待定系數(shù)法,先設(shè)復(fù)數(shù)z=a+bi,再根據(jù)zi=-1+i列出等量關(guān)系,得到a、b的值即可求解.

【仿真訓(xùn)練】一、選擇題1.(3+4i)-(-3-4i)= (

) A.6 B.8i C.6+8i D.0【答案】C2.(3+4i)2= (

) A.-7+24i B.25+24i C.-7 D.25【答案】A

【答案】B4.(1+i)+(2-5i)= (

) A.1-4i B.3-4i C.3+4i D.-1-4i【答案】B5.已知復(fù)數(shù)z滿足z-i=iz+3,則復(fù)數(shù)z= (

) A.-1-2i B.-1+2i C.1-2i D.1+2i【答案】D6.若復(fù)數(shù)(1-i)(a+i)為純虛數(shù),則實數(shù)a= (

) A.2 B.1

C.-1

D.0【答案】C7.已知復(fù)數(shù)z滿足(2-i)z=1+2i,則復(fù)數(shù)z的虛部為

(

) A.1 B.-1

C.0

D.i【答案】A8.若復(fù)數(shù)z滿足z-3i=5+2i,則z= (

) A.5i B.5+5i C.5-i D.5-5i

9.已知z1,z2在復(fù)平面內(nèi)的對應(yīng)點關(guān)于實軸對稱,且z1=1+2i,則z1·z2= (

) A.-5 B.1-2i C.-3 D.5

10.已知a∈R,若(2+ai)(a-2i)=-4i,則a= (

) A.1 B.-1

C.0

D.2

二、填空題11.(2-3i)2=

.

12.若復(fù)數(shù)z滿足z+(5-6i)=3,則z的實部是

,虛部是

.

13.若復(fù)數(shù)z=(2+