12.2復(fù)數(shù)的運(yùn)算第1課時(shí)復(fù)數(shù)的加法、減法與乘法運(yùn)算(教學(xué)方式:基本概念課——逐點(diǎn)理清式教學(xué))[課時(shí)目標(biāo)]1.結(jié)合實(shí)數(shù)的四則運(yùn)算法則,熟練掌握復(fù)數(shù)代數(shù)形式的四則運(yùn)算法則.2.會(huì)應(yīng)用復(fù)數(shù)的四則運(yùn)算法則進(jìn)行復(fù)數(shù)的運(yùn)算.逐點(diǎn)清(一)復(fù)數(shù)的加、減運(yùn)算[多維理解]1.復(fù)數(shù)的加法運(yùn)算及運(yùn)算律(1)復(fù)數(shù)的加法法則設(shè)z1=a+bi,z2=c+di(a,b,c,d∈R),則(a+bi)+(c+di)=.

(2)復(fù)數(shù)加法滿足的運(yùn)算律對(duì)任何z1,z2,z3∈C,有z1+z2=,(z1+z2)+z3=.

2.復(fù)數(shù)的減法運(yùn)算(1)復(fù)數(shù)的差把滿足(c+di)+(x+yi)=a+bi的復(fù)數(shù)x+yi(x,y∈R)叫作復(fù)數(shù)a+bi減去c+di所得的差,記作.

(2)復(fù)數(shù)的減法法則(a+bi)-(c+di)=.[微點(diǎn)練明]1.已知復(fù)數(shù)z1=12-32i和復(fù)數(shù)z2=cos60°+isin60°,則z1+z2等于(A.1B.-1C.12-32iD.122.復(fù)數(shù)z1=a+4i,z2=-3+bi,若它們的和為實(shí)數(shù),差為純虛數(shù),則實(shí)數(shù)a,b的值為()A.a=-3,b=-4 B.a=-3,b=4C.a=3,b=-4 D.a=3,b=43.已知x∈R,y∈R,(xi+x)+(yi+4)=(y-i)-(1-3xi),則x=,y=.4.(1)(i2+i)+(1+i)=.

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

(3)(8-2i)-(-7+5i)+(33+7i)=.

逐點(diǎn)清(二)復(fù)數(shù)的乘法運(yùn)算[多維理解]1.復(fù)數(shù)的乘法法則(a+bi)(c+di)=.

2.復(fù)數(shù)乘法的運(yùn)算律對(duì)任何z1,z2,z3∈C,有交換律z1z2=

結(jié)合律(z1z2)z3=

分配律z1(z2+z3)=

[微點(diǎn)練明]1.計(jì)算:(1-i)2-(2-3i)(2+3i)=()A.2-13i B.13+2iC.13-13i D.-13-2i2.(2022·新課標(biāo)Ⅱ卷)(2+2i)(1-2i)=()A.-2+4i B.-2-4iC.6+2i D.6-2i3.(2023·全國(guó)甲卷)設(shè)a∈R,(a+i)(1-ai)=2,則a=()A.-2 B.-1 C.1 D.24.計(jì)算下列各題.(1)(1-i)(1+i)+(2+i)2;(2)(2-i)(-1+5i)(3-4i)+2i.逐點(diǎn)清(三)共軛復(fù)數(shù)[多維理解]1.共軛復(fù)數(shù)的定義(1)我們把相等、虛部互為的兩個(gè)復(fù)數(shù)叫作互為共軛復(fù)數(shù).

(2)記法:把復(fù)數(shù)z=a+bi的共軛復(fù)數(shù)記作,即z=.

2.共軛復(fù)數(shù)的性質(zhì)當(dāng)復(fù)數(shù)z=a+bi的虛部b=0時(shí),z=z,也就是說(shuō),實(shí)數(shù)的共軛復(fù)數(shù)是它本身.[微點(diǎn)練明]1.復(fù)數(shù)z=1-4i的共軛復(fù)數(shù)是()A.1+4i B.-4+iC.-1+4i D.-1-4i2.(2024·全國(guó)甲卷)設(shè)z=2i,則z·z=()A.-2 B.2 C.-2 D.23.若x-2+yi和3x-i互為共軛復(fù)數(shù),則實(shí)數(shù)x=,y=.

4.復(fù)數(shù)z滿足(1+2i)z=4+3i,則z=.第1課時(shí)復(fù)數(shù)的加法、減法與乘法運(yùn)算[多維理解]1.(1)(a+c)+(b+d)i(2)z2+z1z1+(z2+z3)2.(1)(a+bi)-(c+di)(2)(a-c)+(b-d)i[微點(diǎn)練明]1.選A因?yàn)閦2=cos60°+isin60°=12+32i,所以z1+z22.選A由題意可知z1+z2=(a-3)+(b+4)i是實(shí)數(shù),z1-z2=(a+3)+(4-b)i是純虛數(shù),故b+4=0,a+3=0,4-b≠0,解得a=-33.解析:由已知得,x+4+(x+y)i=(y-1)+(3x-1)i,∴x+4=y答案:6114.解析:(1)原式=(-1+i)+(1+i)=-1+i+1+i=2i.(2)原式=(-1+3i)+(-2-i)+(1-2i)=(-3+2i)+(1-2i)=-2.(3)原式=[8-(-7)+33]+(-2-5+7)i=15+33.答案:(1)2i(2)-2(3)15+33[多維理解]1.(ac-bd)+(bc+ad)i2.z2z1z1(z2z3)z1z2+z1z3[微點(diǎn)練明]1.選D(1-i)2-(2-3i)(2+3i)=1-2i+i2-(4-9i2)=-13-2i.2.選D(2+2i)(1-2i)=2-4i+2i+4=6-2i.3.選C∵(a+i)(1-ai)=a+i-a2i-ai2=2a+(1-a2)i=2,∴2a=2且1-a2=0,解得a=1,故選C.4.解:(1)(1-i)(1+i)+(2+i)2=1-i2+4+4i+i2=5+4i.(2)(2-i)(-1+5i)(3-4i)+2i=(-2+10i+i-5i2)(3-4i)+2i=(3+11i)(3-4i)+2i=(9-12i+33i-44i2)+2i=53+21i+2i=53+23i.[多維理解]1.(1)實(shí)部相反數(shù)(2)za-bi[微點(diǎn)練明]1.A2.選D因?yàn)閦=2i,所以z=-2i,z·z=2,故選D.3.解析:∵x-2+yi和3x-i互為共軛復(fù)數(shù),所以x-2=3x,且y=1,所以x=-