Thiết kế bài giảng giải tích 12 (tập 2) phần 2

ChirONq IV

SO PHLTC
Phan 1
msta^G

TAX D E CUA CHUUlVG

I. NOI DUNG
Ndi dung chinh cua chuang 4 :
- So phiic : Dinh nghia ; hai sd phiic bdng nhau; bieu dien hinh hgc ciia so phiic;
md dun ciia sd phiic T sd phiic lien hgp.
Cac phep toan ve sd phiic : Phep cdng va phep trii; phep nhan cac sd phiic ;
Tong va tfch hai sd phiic lien hgp ; phep chia hai sd phiic.
Phuong trinh bac hai ddi vdi he sd thuc : Can bac hai ciia sd thuc am; phuong
trinh bac hai ddi vdi he so thuc.
n . MUC TIEU
1. Kien thurc
Ndm dugc toan bd kien thiic co ban trong chuang da neu tren, cu the :
Ndm viing dinh nghia so phiic va cac phep toan ciia nd.
• Hieu dugc mddun cua so phiic va bieu dien mdi sd phiic tren mat phang tga do.
Mdi quan he ciia hai so phiic lien hgp.
2. KT nang.
Van dung thanh thao cac phep toan.
• Tim dugc mddun cua mdt so phiic.
• Giai dugc phuang trinh bac hai cd nghiem phiic.

77


3. Thai do

Tur giac, tich cue, dgc lap va chii dgng phat hien ciing nhu linh hdi kien thirc
trong qua trinh hoat dgng.
Cam nhan dugc su cdn thie't eiia dao ham trong viec khao sat ham sd.
Cam nhan duoc thuc te' ciia toan hgc, nha't la ddi vdi dao ham.

78


Phan 2
CAC B A I HOAX

§1. So' phu'c
(tiet 1, 2, 3)
I. MUC TIEU
1. Kien thurc
HS ndm dugc :
So i la gi? Y nghia ciia nd.
• Dinh nghia so phiic
Hai so phiic bdng nhau khi nao?
Bieu dien hinh hgc sd phiic.
Mddun ciia sd phiic.
2. KT nang
HS tfnh mddun cua so phiic.
• Tfnh thanh thao so phiic lien hgp cua mdt sd phiie.
3. Thai do
- Tu giac, tich cue trong hgc tap.
• Biet phan biet rd cac khai niem co ban va van dung trong tung trudng hgp cu the.
- Tu duy cac van de cua toan hgc mdt each Idgic va he thdng.
n . CHUAN BI CUA GV VA HS
1. Chuan hi cua GV

• Chudn bi cac cau hdi ggi md.
• Chudn bi phan mau, va mdt so dd diing khac.
2. Chuan bj cua HS
Cdn dn lai mdt sd kien thiie da hgc phuong trinh bac hai.
79


m . PHAN PHOI THCJI LUONG
Bai nay chia lam 3 tiet:
Tii't I : Tit ddu den hit muc 3.
Tii't 2 : Tii'p theo din hit muc 5.
Tiet 3 : Tii'p theo den hit muc 6 vd hudng ddn bdi tap..
IV. TIEN TRINH DAY

HOC

A. OAT VAN OE
Cau hdi 1
Xet tinh diing - sai ciia cac cau sau day :
a) Cd sd thuc x ma .x^ = 1.
b) Cd sd thuc x ma x = - 1 .
Cau hdi 2
Chiing minh phuong trinh sau khdng cd nghiem thuc
a)x^-2x + 3.

b)-x^+x-7

GV:
Sd i thda man i^ = -1 ta ggi dd la mgt sd phiic.

B. BAI Mdl
HOAT DONG 1
1. Sd i
HI. Cd nhiing sd am nao khi binh phuang thi bdng 1.
H2. Cd nhiing sd am nao khi binh phuang thi bdng - 1 .
H3. Phai chang cd mdt so khdng la so phiic ma khi binh phiuong bang - 1 ,
• GV neu khai niem' sd i:
Nghiem cua phuang trinh x +1 = 0 Idso i.
2

Nhu vay : i = - 1 .
80


HOAT DONG 2
2. Djnh nghTa so phiic
• GV neu dinh nghia sd phiic:
Mdi bieu thdc dgng a + bi; a, b e R i = - 1 duac ggi la mot so phicc.
Ddi vdi sd phicc z = a + bi, ta noi a la phdn thuc, b la phdn do cua z.
Tap hgp cdc sdphdc ki hieu Id C
C = {a + 6i I a, 6 e R, i^ = - l )
H4. Hay neu vi du ve so phutc.
H5. Sd thuc la trudng hgp rieng ciia sd phiic. Diing hay sai?
• Thuc hien .QL 1 trong 5'
Hoat dgng ciia GV
Cau hdi 1

Hoat dgng ciia HS
Ggi y tra idi cau hdi 1

Tim phan thuc va phdn ao ciia GV ggi mdt vai HS tra Idi.
so phiic - 3 + 5i
Phdn thuc :,,,.
Phdn ao :,,,.
Sau dd ket luan.
Cau hdi 2

Ggi y tra Idi cau hdi 2

Tim phdn thuc va phdn ao ciia GV ggi mdt vai HS tra Idi.
Phdn thuc :....
so phiic 0 + 7ti.
Phan ao :....
Sau dd ket luan.
Cau hdi 3

Ggi y tra Idi cau hdi 3

Tim phdn thuc va phan ao ciia GV ggi mdt vai HS tra Idi.
so phiic 1 + Gi
Phdn thuc :....
Phdn ao :....
Sau dd ket luan.

H6. Phai chang ca phdn thuc va phdn ao cua mdt so phiic la mdt so thuc?

81


HOAT DONG 3

3. Hai sd phurc b^ng nhau
• GV neu dinh nghia :
Hai sdphitc Id bdng nhau neu phdn thuc vd phdn do cua chung tuang
dng bdng nhau.
a + bi = c + di<::>a = c v a b = d
H7. Hay neu mdt so vi du ve hai so phiic bdng nhau.
H8. Cho sd phiic : V2 + 3i. Sd nao sau day bdng sd phiic tren
(a)2-V3i;
(a)

(a)-2-V3i

4 + 2V3i

(a) 4 + V3i

• Thuc hien vi du 2 trong 5' GV cd th^ thuc hien vi du khac.
Hoat dgng ciia GV
Cau hdi 1

Hoat dgng cua HS
Ggi y tra Idi cau hdi 1

Mdi quan hd ciia x va y de hai
sd phiic dd bang nhau.

2x + l = x + 2 v a 3 y - 2 = 3/ + 4.
Ggi y tra Idi cau hdi 2

Cau hdi 2

Tim X va y

HS giai he tren ta cd : x = 1 va y = 3.

H9. Tim cac so thuc x va y, biet
(x + 1) + iSy - 2)1 = i-x + 2) + (2y + 4)i.
HIO. Tim cac so thuc x vay, biet
(-X + 1) + (2y - 1)1 - (x + 2) + (y + 4)1.
• GV neu chii y :
• Mdi sd thuc a dugc coi la mdt sd phiic vdi phdn do bdng 0
a = a + Oi.
Nhu vay, mdi so thuc cung la mot sophHtc. Ta cd R c C.
• Sd phiic 0 + bi dugc ggi la sd do va vie't dan gian la bi
bi = 0 + bi.
82


Dac biet
i = 0 + li.
So / dugc ggi la dan vi do.
Hll. Hay chi ra phdn thuc va phdn do cua cac sd sau:
a) 7 ;

b) -4i

Sd nao la sd thudn do?
• Thuc hien ^ ; 2 ti'ong 5'.
Hoat dgng ciia HS

Hoat dgng ciia GV

Cau hdi 1

Ggi y tra Idi cau hdi 1

Hay viet so phiic z thda man
dd bai.

1
z=

^ .
1.

2 2
Ggi y tra Idi cau hdi 2

Cau hdi 2

Hay viet sd phiic z cd phdn ao GV ggi mdt vai HS tra Idi.
5, phdn thuc V2 .

z = yj2 + 5i.

Cau hdi 3

Ggi y tra Idi cau hdi 3

Hay vie't sd phurc z cd phdn ao z= V 2 - 5 i .
- 5 , phdn thuc V2
HOAT DONG 4

4. Bieu di^n hinh hgc ciia sd phirc
• GV neu dinh nghia :
Diem M(a ; b) trong mot he tog do vuong goc cua mat phdng duac ggi
la diem bieu diin sd phutc z = a + bi
• GV sir dung hinh 68 de dat cac cau hdi:
H12. Bieu dien cac so phiic sau tren mat phdng:
a ) l + 3i;

b ) 2 + V3i;

c) l - 3 i ;

c) 2-V3i.
83


HI3. Tim tap hgp cac sd phiic tren mat phang tga do chi cd phdn ao.
HI4. Tim tap hgp cac sd phiic tren mat phang tga do chi cd phdn thuc.
HI5. Hai sd phiic dugc bieu diln tren mat phang tga do cd dac diem gi ne'u:
a) Cd phdn thuc bang nhau nhung phdn ao ddi nhau.
b) Cd phdn do bdng nhau nhung phdn thuc ddi nhau.
c) Cd phan thuc va phdn do ddi nhau.
• Thuc hien -^

3 trong 5'
Hoat dgng cua HS

Hoat dgng cua GV

Ggi y tra Idi cau hdi 1

Cau hdi 1

HS tu bieu didn

Bieu dien so phiic 3 - 2i

Ggi y tra Idi cau hdi 2

Cau hdi 2

HS tu lam.

Bieu dien so phiic - 4 i , 3.
Cau hdi 3

Ggi y tra Idi cau hdi 3

Tra Idi cau b.

HS tu lam.
HOAT DONG 5

5. Md dun cua sd phiic
• GV neu dinh nghia :
.
I,

o

M

II

X

Do ddi cua vecta OM duac ggi la mddun cua sd phicc z vd ki hieu la jzj.
Vay \a + bi\ = Va^ + b^
84


• Thuc hien ^ 4 trong 5'
Hoat dgng cua HS

Hoat dgng ciia GV
Cau hdi 1

Ggi y tra Idi cau hdi 1

Tim so phiic cd mo dun bdng 0.

Va^ + &^ = 0 » a = & = 0.
SdO.
Ggi y tra Idi cau hdi 2

Cau hdi 2
Tim so phiic cd mddun bang
1.

Va^ + 6^ = 1 « a^ + 6^ = 1.

Cac so z = 1, z = i, z = -i,... cd mddun
bdng 1.

H16. Mdi sd phiic deu cd mdt mddun. Diing hay sai?
H17. Hai sd phiic bdng nhau cd mddun bdng nhau. Diing hay sai?
HI8. Hai so phirc cd mddun bdng nhau thi bdng nhau. Diing hay sai?
HOAT DONG 6
6. Sd phiic lien hgp
• Thuc hien ^^ 5 trong 5'
Hoat dgng cua GV
Cau hdi 1
Bieu dien hai sd :

Hoat dgng cua HS
Ggi y tra Idi cau hdi 1
HS tu bieu diln.

2+ 3i va 2 - 3i tren mat phang Hai diem nay ddi xiing qua Ox.
tga do.
Ggi y tra Idi cau hdi 2
Cau hdi 2
HS tu bieu diln.
Bieu diln hai sd :
— 2 + 3i va - 2 — 3i tren matHai diem nay ddi xiing qua Ox.
phang tga do..

85


• GV neu dinh nghia :

Cho sd phicc z = a + bi. Ta ggi a - bi la sd phiic liin hop cua z vd
ki hieu la 'z - a - bi.
H19. Hai sd phiic lien hgp cd ciing mddun. Dung hay sai.
H20. Hay neu phan vi du ve hai sd phiic cd ciing mddun nhung khdng phai hai sd
phiic lien hgp.
• Thuc hien .Ql 6 trong 5'
Hoat dgng ciia HS

Hoat dgng ciia GV

Ggi y tra Idi cau hdi 1

Cau hdi 1

z = 3-2i.

Tim z.

Ggi y tra Idi cau hdi 2

Cau hdi 2

z = z = 3 - 2i.

Tim z.

Ggi y tra Idi cau hdi 3

Cau hdi 3
Tfnh |z| va z

z = |z| = Vl3

• GV neu ket luan:
z = z.
\z\ = \z\.
• GV neu va thuc hien vi du 4. GV cd the thay bdi vi du khac.
H21. Tim sd phiic lien hgp ciia z = 13 -5i.
H22. Tim sd phiic lien hgp cua z = -13 -5i.
H23. Tim sd phiic lien hcrp ciia z = 13 +5i.
H24. Tim sd phiic lien hgp ciia z = -13 + 5i.

86


HOATTX:>NG 7

TOM TflT Bfll HQC
1. Mdi bieu thiic dang a + &/; a, 6 e R

i = - 1 dugc ggi la mdt sd phurc.

Ddi vdi so phiic z = a + bi, ta ndi a la ph^n thuc, b la phan ao ciia z.
Tap hgp cac sd phiic kf hieu la C.
C = {a + 6i |a, 6 e R, i^ = - l } 2. Hai sd phiic la bang nhau ne'u phdn thuc va phdn ao ciia chiing tuong ling bdng
nhau. a + 6i = c + (ii<=>a = c v a 6 = (i.
3. Dilm Mia ; 6) trong mdt he toa do vudng gdc cua mat phang dugc ggi la diem
bieudi^nsophiirc z-a + bi
4. Dd dai ciia vecta OM dugc ggi la mddun ciia so phiic z va ki hieu la \z\.
\a + bi\ - V a ^ + 6 ^

5. Cho sd phiic z = a + bi.Ta ggi a - bi la so phurc lien hgp ciia z va kf hieu la
z - a - bi.
* z = z.
-\z\ = \zl
HOAT DONG 8

MQT SO CflU H 6 | TRflC NGHIEM ON TflP Bfll 1
Dien diing sai v^o chd trdng sau:
Cdu 1. Cho sd phiic z = 2 - 5i

(a)z=2 + 5i

•

(b)|z| = V29

•
87


(c) 2:

D
D

-V29

(d) z = z
Trd Idi.
a

b

c

d

D

D

D

S

Cdu 2. Cho sd phiic z = 2 + 5i

•

(a)z = 2 - 5 i

D
D
D

(b) |z| = V29
(c) z =V29
(d) z = z
Trd Idi.
a

b

c

d

D

D

D

S

Cd i 3. Cho sd phiic z = 2 + 5i, z' = a + bi
(a) z khdng the bdng z'

D

fa = 2
(b)z = z ' « | ^ ^ ^

D

fa-2

D

(c)z = z ' «

[b = —5

fa = 2
(d)z'=ZC:>
[b = —5

Trd Idi.

88

a

b

c

d

S

D

S

D

•

Cdu 4. Cho hai sd phiic z = 2 - yi, z' = 5x + 3i; z = z' khi
2

, y = -3 ;

(a) X = 2, y = -3 ;

(b)x

(c) X = -2, y = -3 ;

(d)x = 2,y = 3 ;

Trd Idi. (b).
Cdu 5. Cho hai so phiic z = 2 - yi, z' = 5x + 3i; z = z' khi
(a) X = 2, y = -3 ;

(b)x = - - , y = 3 ;

(c) X = -2, y = -3 ;

(d)x = - - , y = - 3 ;

Trd Idi. (d).
Cdu 6. Cho sd phiic z = 12 - v3 i. Sd z la :
(a) i = 12 + V3 i;

(b) z = 12 - Vs i;

(c) z = -12 - V3 i;

(d) z = -12 + Vs i.

Trd Idi. (a).
Cdu 7. Cho sd phiic z = 12 - v3 i; Izl bang
(a) V147 ;

(b) V15

(c)^/21;

(d) Vl53

Trd Idi. (a).
Cdu 8. Cho so phiic z = vl2 - v3 i; |z| bdng
(a) V147 ;

(b) Vl5

(c) V2T;

(d) V153

Trd Idi. (b).

89


Cdu 9. Cho sd phiic z = vl2 - 3 i; |z| bang
(a) 7147;

(b) Vis

(c) V2I;

(d) Vl53

Trd Idi: ic).
: 12 - 3 i; |z| bang
au 10. Cho sd phurc z =
(b) Vl5

(a) VT47 ;

(d) VTsI

(c)V21;
Trd Idi. (d).

HOAT DONG 9

md^^ DflN Bfll TflP SflCH GIflO KMOfl
Bai 1. Hudng ddn. z = a + bi thi phdn thuc a, phdn do b.
GV cho HS len bang dien vao d trdng
a)
Phdn thuc

Phdnao

1

Tt

Phdn thuc

Phdnao

b)

V2,

-1 '

c)
Phdn thuc

Phdnao

2V2,

0

90


d)
Phdn thuc

Phdn ao

0

-7

Bai 2. Hudng ddn. Six dung cac tfnh chdt ciia hai sd phiic bang nhau
cau a. Giai he

f3x-2 = x + l
2y + l = - ( y - 5 )

r>' -'
3
4
Dap so. X-—, y = — .
2
3
cau b. Giai he :

Ddp so. x =

[l - 2x = VS
[l - 3y = S

1-S
2

cau c. Giai he

'^

l + yfS
3

r2x + y = x - 2 y + 3
[2y - x = y + 2x + l

Ddpsd. X = 0, y = 1.
B^i 3. Hudng ddn. Six dung cac tinh chat ciia sd phiic
cau a. Thudc dudng thdng x = -2.
Ddp sd. Hinh ve.

y^

-2

O

91


cau b. Thudc dudng thang y = 3
Ddp sd. Hinh ve.

^

cau c. La phdn dugc gidi han bdi hai dudng thang x = -1 va x = 2.
Ddp sd. Hinh ve.

y|

!2

-IK;Q

X

cau d. La phdn dugc gidi han bdi hai dudng thang y = 1 va y = 3.
Ddp sd. Hinh ve.
y-

WM/kwm.
3

1

o

X

cau e. La phdn dugc gidi han bdi hinh vudng.
Ddp sd. Hinh ve.

m
92

2

X

Bai 4. Hudng ddn. Six dung cac tfnh chdt ciia mddun sd phiic.
Neu z = a + ib thi |z| = Va^ + b^
cau a. Ddp sd. I z I = 'ji-2f+iSf

= 4l

Cau b. Ddp sd \z\ = ^[(Sf^+i^
cau c. Dap sd. \ z \ = \]i-5)
cau d. Ddp sd. I z_| = ^[iSf

= Vn

- 5
= V3.

Bai 5. Hudng ddn. Tinh ehdt ciia mddun so phiic tren mat phang tga do.
Cau a. Ddp sd. La dudng trdn ban kfnh 1.

cau b. Dap sd. La hinh trdn ban kfnh 1.
y|

93


cau c. Dap sd. La phdn hinh gidi han bdi hai hinh trdn ban kinh 1 va 2.
y*

caud.

Ddp sd. La dudng trdn ban kinh 1.

Bai 6. Hudng ddn. Dua vao dinh nghia sd phiic lidn hgp.
cau a. Ddp sd. z =1 + iyf2 .
cau b. Ddp sd. z = -V2 - iy/s
cau a. Ddp sd. z = 5.
Cau a. Ddp sd. z = -71.

94


§2. C o n g tru* v a n h a n so' phuTc
(tiet 4, 5)
I. MUC T l £ u
1. Kien thurc
HS ndm dugc :
Khai niem phep cdng, phep trii so phiie.
- Dinh nghia phep cdng sd phiic.
- Dinh nghia phep trir sd phiic.
Phep nhan so phiic.
2. KT nang
Van dung thanh thao cac phep toan cdng va trii so phiic.
Ket hgp cac tinh chdt de thuc hien eac phep toan.
So sanh vdi cac phep toan ciia sd Ihuc.
3. Thai do
- Tu giac, tich cue trong hgc tap.
Biei phan biet rd cac khai niem co ban va van dung trong tumg trudng hgp cu thi.
- Tu duy cac vdn de ciia toan hgc mdt each Idgic va he thdng.
n . CHUAN BI CUA GV VA HS
1. Chuin bj cua GV
Chudn bi eac cau hdi ggi md.
Chudn bi pha'n mau va mdt sd dd diing khac.

2. Chuan bj cua HS
Cdn dn lai mdt sd kie'n thiic da hgc bai 1.
m.

PHAN

PH6I

THCJI LUONG

Bai nay chia lam 2 tie't:

95


Tiit I : Tic ddu din hit phdn 1.
Tie't 2 : Tii'p theo din hit phdn 2.
IV. TIEN TRINH DAY

HOC

A. OAT VAN OE
Cau hdi 1
Neu cac khai niem vl sd phiic :
- Dinh nghia sd phiic.
- Sd phiic lien hgp.
- Mddun cua so phiic.
- Bieu diln hinh hgc eiia sd phiic.
Cau hdi 2
Tim so phiic lien hgp cua cac sd sau:

a)z = 4 - 7 i ;
Cau hdi 3

b)z=V3-5i

Tim mddun cia cac so phiic sau :
a)z = 4 - + 7 i ;

b)z=V3+5i

B. BAI Mdl
HOAT DONG 1
1. Phep cdng va phep trur
• Thuc hien . ^ 1 trong 5'

cau 1)
Hoat dgng ciia GV

Hoat dgng ciia HS

Cau hdi 1
Tinh (3 + 2i) + (5 + 8i)

Ggi y tra Idi cau hdi 1
(3 + 2i) + (5 + 8i) = 8 + lOi

Cau hdi 2
Tinh (7 + 5 i ) - ( 4 + 3i)

Ggi y tra Idi cau hdi 2

(7 + 5 i ) - ( 4 + 3 i ) - 3 + 2i

96


Ggi y tra Idi cau hdi 3

Cau hdi 3
Tfnh (-V3 + 5 i ) - ( 4 - 3 i )

(-VJ + 5i) - (4 - 3i) = (-V3 - 4) + 2i

• GV neu dinh nghia
Phep cong vd phep triJC hai sdphCcc duac thUc hien theo quy tac cong,
trie da thdc.
HI. Tinh (7 + 5i) + (4 + 3i).
H2.TInh (7 + 5 i ) - ( 4 + 3i).
H3.Tfnh -(7 + 5 i ) - ( 4 + 3i).
H4.TInh(7 + 5 i ) - ( 4 - 3 i ) .
• Thuc hien vi du 1 trong 5'

Hoat dgng ciia GV
Cau hdi 1
Tfnh (5 + 20 + (3 + li)

Hoat dgng cua HS
Ggi y tra Idi cau hdi 1
HS tu tinh.

Cau hdi 2

Tfnh (1 + 6i) - (4 + 3 0 .

Ggi y tra Idi cau hdi 2
HS tu tinh.

' GV neu tong quat:
(a + bi) + (c + di) = (a + c) + (b + d)i ;
(a + bi) - (c + di) = (a - c) + (b - d)i.
H6. Tim so phiic lien hgp ciia z-

(2 - 3i) - (5 + 4i).

H7. Tun sd phiic lien hgp ciia z-

(2 - 3i) + (5 + 4i)

H8. Tim sd phiic lien hgp ciia z = (2 - 3i) - (5 - 4i)
H9. Tim sd phiic lien hgp cua z = (2 + 3i) - (5 + 4i)

97


HOAT DONG 2
2. Phep nhan
• Thuc hien ^ ^ 2 trong 5'
Hoat dgng ciia HS

Hoat dgng cua GV

Ggi y tra Idi cau hdi 1

Cau hdi 1
Tinh (3 + 2i)(2 + 3i).
Cau hdi 2

(3 + 2i)(2 + 3i) = 6 + 13i + 6i2 = 13i.
Ggi y tra Idi cau hdi 2

Tinh (3 + 2i)(2-3i).

(3 + 2i)(2-3i) = 6 - 5 i - 6 i 2 = 1 2 - 5 i .

• GV neu dinh nghia
Phep nhdn hai sdphdc duac thuc hien theo quy tdc nhdn da thicc roi
•2

thay i - -1 trong kit qua nhdn duac.
- Thuc hien vi du 2 trong 5'. GV cd the Idy vi du khac
Hoat dgng ciia HS

Hoat dgng cua GV

Ggi y tra Idi cau hdi 1

Cau hdi 1
Tinh (5 + 20(4 + 30

HS tu tfnh
Ggi y tra Idi cau hdi 2

Cau hdi 2
Tfnh (5 + 20(4 + 3 0 .

HS tu tfnh.

HIO. Tim sd phiic lidn hgp ciia z = (2 - 3i)(5 + 4i).
Hll. Tim so phiic lien hgp ciia z = (2-3i)(5 + 4i)
HI2. Tim sd phiic hen hgp ciia z = (2 - 3i)(5 - 4i)
H13. Tim sd phiic lien hgp cua z- (2 + 3i)(5 + 4i)
• GV neu tong quat:
(a + bi)ic + d o = (ac - bd) + iad + bc)i.

98


HOAT DONG 3

TdM TflT Bfll HPC
1. Phep cdng va phep trir hai sd phiic dugc thuc hien theo quy tdc cdng, trir da thiic.
(a + 60 + (c + di) = (a + c) + (6 + d)i ;
(a + 60 - (c + di) = (a - c) + (6 - d)i.
2. Phep nhan hai sd phiic dugc thuc hien theo quy tdc nhan da thiic roi thay
O

i = -1 trong ket qua nhan dugc.
(a + bi)ic + di) = iac - bd) + iad + bc)i.
HOAT DONG 4

MQT SO CflU HOI TR^C NGHIEM KHflCH QUflN
Hdy dien dung sai vao 6 trdng sau:

Cdu 7. Cho z = 2 + 4i, z' = 5 - 3i
(a) z la so phiic lien hgp cua z'
(b) z + z' = 7 + i
(c) z - z' = -3 + 7i
(d) zz' = 22 + 14i

0
D
D
D

Trd Idi
(a) (b)

(c)

(d)

S

D

D

D

Cdu 2. Cho z = 2 + 4i, z' la so phirc lien hgp ciia z
(a) z' = 2 - 4i

D

99


D
D
D

(b) z + z' = 0
(c) z - z' = 8i
(d) zz' = 20
Trd Idi
(a) (b)
D

S

(c)

(d)

D

D

Cdu 3. Cho z = 2 + V3 i, z' la sd phiic lien hgp ciia z

(b) z + z' = 0

D
D

(c) z - z' = 0

•

(d) zz' = 7

D

(a)z' = 2-V3i

Trd Idi
(a) (b)

(c)

(d)

D

S

D

S

Cdu 4. Cho z- v3 i, z' la so phiic lien hgp ciia z

D
D

(a)z'= -V3i
(b) z + z' = 0

•

(c) z - z' = 0
(d) zz' = 3
T?-d Idi

100

(a) (b)

(c)

(d)

D

S

D

D


Hdy chgn khdng dinh dung trong cdc cdu sau:
Cdu 5. Cho z = (3 + 2i) + (5 - i). So phiic lien hgp cua z la
(a) 8 + i;

(c) -8 - i;

(b) 8 - i
(d) -8 + i.

Trd Idi. (a).
Cdu 6.. Cho z = (3 + 2i) - (5 - i). Sd phiic lien hgp cua z la
(a) -2 + 3i;
(c) 2+3i;

(b) -2 - 3i
(d) 2 - 3i.

Trd Idi. (a).
Cdu 7. Cho z = (3 + 2i) - (5 - i); z' = i.
(a) zz' = 3 - 2 i ;
(c) -2+3i;

(b)-2i - 3
(d)2 + 3i.

Trd Idi. (a).
Cdu S. Cho z = (3 + 2i) + (5 --i); z ' = i.
(a)zz'=-l+8i »
(c) 1 + 8 i ;

(b) - 1 --8i
(d)i.

Trd Idi. (a).

HOAT DONG 5

HaCTNG DflN Bfll TflP SGK
Bai 1. Hudng ddn. Sur dung dinh nghia phep cdng, phep trii sd phiic.
Cau a. Ddp sd. 5 - i.
cau b. Ddp sd.

-3-lOi.

cau c. Ddp sd. - 1 + lOi
Cau d. Ddp sd. -3 + i.

101