ChirONq IV SO PHLTC Phan msta^G TAX D E CUA CHUUlVG I NOI DUNG Ndi dung chinh cua chuang : - 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 Kien thurc Ndm dugc toan bd kien thiic co ban 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 Mdi quan he ciia hai so phiic lien hgp KT nang Van dung thao cac phep toan • Tim dugc mddun cua mdt so phiic • Giai dugc phuang trinh bac hai cd nghiem phiic 77 Thai Tur giac, tich cue, dgc lap va chii dgng phat hien ciing nhu linh hdi kien thirc qua trinh hoat dgng Cam nhan dugc su cdn thie't eiia dao ham viec khao sat ham sd Cam nhan duoc thuc te' ciia toan hgc, nha't la ddi vdi dao ham 78 Phan CAC B A I HOAX §1 So' phu'c (tiet 1, 2, 3) I MUC TIEU Kien thurc HS ndm dugc : So i la gi? Y nghia ciia nd • Dinh nghia so phiic Hai so phiic bdng nao? Bieu dien hinh hgc sd phiic Mddun ciia sd phiic KT nang HS tfnh mddun cua so phiic • Tfnh thao so phiic lien hgp cua mdt sd phiie Thai - Tu giac, tich cue hgc tap • Biet phan biet rd cac khai niem co ban va van dung tung trudng hgp cu the - Tu cac van de cua toan hgc mdt each Idgic va he thdng n CHUAN BI CUA GV VA HS Chuan hi cua GV • Chudn bi cac cau hdi ggi md • Chudn bi phan mau, va mdt so dd diing khac Chuan bj cua HS Cdn dn lai mdt sd kien thiie da hgc phuong trinh bac hai 79 m PHAN PHOI THCJI LUONG Bai chia lam tiet: Tii't I : Tit ddu den hit muc Tii't : Tii'p theo din hit muc Tiet : Tii'p theo den hit muc vd hudng ddn bdi tap IV TIEN TRINH DAY HOC A OAT VAN OE Cau hdi Xet tinh diing - sai ciia cac cau sau day : a) Cd sd thuc x ma x^ = b) Cd sd thuc x ma x = - Cau hdi Chiing minh phuong trinh sau khdng cd nghiem thuc a)x^-2x + b)-x^+x-7 GV: Sd i thda man i^ = -1 ta ggi dd la mgt sd phiic B BAI Mdl HOAT DONG 1 Sd i HI Cd nhiing sd am nao binh phuang thi bdng H2 Cd nhiing sd am nao binh phuang thi bdng - H3 Phai chang cd mdt so khdng la so phiic ma binh phiuong bang - , • GV neu khai niem' sd i: Nghiem cua phuang trinh x +1 = Idso i Nhu vay : i = - 80 HOAT DONG 2 Djnh nghTa so phiic • GV neu dinh nghia sd phiic: Mdi bieu thdc dgng a + bi; a, b e R i = - duac ggi la mot so phicc Ddi vdi sd phicc z = a + bi, ta noi a la phdn thuc, b la phdn cua z Tap hgp cdc sdphdc ki hieu Id C C = {a + 6i I a, 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 5' Hoat dgng ciia GV Cau hdi Hoat dgng ciia HS Ggi y tra idi cau hdi Tim phan thuc va phdn ao ciia GV ggi mdt vai HS tra Idi so phiic - + 5i Phdn thuc :,,, Phdn ao :,,, Sau dd ket luan Cau hdi Ggi y tra Idi cau hdi Tim phdn thuc va phdn ao ciia GV ggi mdt vai HS tra Idi Phdn thuc : so phiic + 7ti Phan ao : Sau dd ket luan Cau hdi Ggi y tra Idi cau hdi Tim phdn thuc va phan ao ciia GV ggi mdt vai HS tra Idi so phiic + 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 Hai sd phurc b^ng • GV neu dinh nghia : Hai sdphitc Id bdng neu phdn thuc vd phdn cua chung tuang dng bdng a + bi = c + dia = c v a b = d H7 Hay neu mdt so vi du ve hai so phiic bdng H8 Cho sd phiic : V2 + 3i Sd nao sau day bdng sd phiic tren (a)2-V3i; (a) (a)-2-V3i + 2V3i (a) + V3i • Thuc hien vi du 5' GV cd th^ thuc hien vi du khac Hoat dgng ciia GV Cau hdi Hoat dgng cua HS Ggi y tra Idi cau hdi Mdi quan hd ciia x va y de hai sd phiic dd bang 2x + l = x + v a y - = 3/ + Ggi y tra Idi cau hdi Cau hdi Tim X va y HS giai he tren ta cd : x = va y = 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 bdng a = a + Oi Nhu vay, mdi so thuc cung la mot sophHtc Ta cd R c C • Sd phiic + bi dugc ggi la sd va vie't dan gian la bi bi = + bi 82 Dac biet i = + li So / dugc ggi la dan vi Hll Hay chi phdn thuc va phdn cua cac sd sau: a) ; b) -4i Sd nao la sd thudn do? • Thuc hien ^ ; ti'ong 5' Hoat dgng ciia HS Hoat dgng ciia GV Cau hdi Ggi y tra Idi cau hdi Hay viet so phiic z thda man dd bai z= ^ 2 Ggi y tra Idi cau hdi Cau hdi Hay viet sd phiic z cd phdn ao GV ggi mdt vai HS tra Idi 5, phdn thuc V2 z = yj2 + 5i Cau hdi Ggi y tra Idi cau hdi Hay vie't sd phurc z cd phdn ao z= V - i - , phdn thuc V2 HOAT DONG 4 Bieu di^n hinh hgc ciia sd phirc • GV neu dinh nghia : Diem M(a ; b) mot he tog 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 ) + V3i; c) l - i ; c) 2-V3i 83 HI3 Tim tap hgp cac sd phiic tren mat phang tga chi cd phdn ao HI4 Tim tap hgp cac sd phiic tren mat phang tga chi cd phdn thuc HI5 Hai sd phiic dugc bieu diln tren mat phang tga cd dac diem gi ne'u: a) Cd phdn thuc bang nhung phdn ao ddi b) Cd phdn bdng nhung phdn thuc ddi c) Cd phan thuc va phdn ddi • Thuc hien -^ 5' Hoat dgng cua HS Hoat dgng cua GV Ggi y tra Idi cau hdi Cau hdi HS tu bieu didn Bieu dien so phiic - 2i Ggi y tra Idi cau hdi Cau hdi HS tu lam Bieu dien so phiic - i , Cau hdi Ggi y tra Idi cau hdi Tra Idi cau b HS tu lam HOAT DONG 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 ^ 5' Hoat dgng cua HS Hoat dgng ciia GV Cau hdi Ggi y tra Idi cau hdi Tim so phiic cd mo dun bdng Va^ + &^ = » a = & = SdO Ggi y tra Idi cau hdi Cau hdi Tim so phiic cd mddun bang Va^ + 6^ = « a^ + 6^ = Cac so z = 1, z = i, z = -i, cd mddun bdng H16 Mdi sd phiic deu cd mdt mddun Diing hay sai? H17 Hai sd phiic bdng cd mddun bdng Diing hay sai? HI8 Hai so phirc cd mddun bdng thi bdng Diing hay sai? HOAT DONG 6 Sd phiic lien hgp • Thuc hien ^^ 5' Hoat dgng cua GV Cau hdi Bieu dien hai sd : Hoat dgng cua HS Ggi y tra Idi cau hdi HS tu bieu diln 2+ 3i va - 3i tren mat phang Hai diem ddi xiing qua Ox tga Ggi y tra Idi cau hdi Cau hdi HS tu bieu diln Bieu diln hai sd : — + 3i va - — 3i tren matHai diem ddi xiing qua Ox phang tga 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 5' Hoat dgng ciia HS Hoat dgng ciia GV Ggi y tra Idi cau hdi Cau hdi z = 3-2i Tim z Ggi y tra Idi cau hdi Cau hdi z = z = - 2i Tim z Ggi y tra Idi cau hdi Cau hdi Tfnh |z| va z z = |z| = Vl3 • GV neu ket luan: z = z \z\ = \z\ • GV neu va thuc hien vi du 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 cau d Hudng ddn Dat u = 2x + ,dvDdpsd e~^dx 3e - Bai 12 Hudng ddn Six dung cac phuang phap ddi bien sd - 71 7t cau a Hudng ddn Dat u - cos(— - 4x) ^> du = 4sin(— - 4x)dx Khi x = o o ^ n _ V3 thi u = — Khi X = TT— thi M = - - - 24 Ddp sd -5- In o ;;— Khi x = - ^ thi t = — cau b Hudng ddn Dat x = —tan^ ^> dx = dcos^^ rr • , , , 71 Khi X = — thi ^ = — - cau c Hudng ddn Dat u = cosx => du = - s i n x d x Khi x = thi u = Khi X = — thi M = z Ddp so — 35 Cau d Hudng ddn Dat u = Vl + t a n x hay u = + t a n x => 2udu = — cos X Khi X = - V thi u = Khi X = ^ thi u ^ S 4 n - so•' 4V2 Dap 147 Bai 13 Hudng ddn Six dung cac phuang phap tinh dien tich hinh phdng b b Chii y den cac cdng thiic : S = j|/(x)|(ic, S = ^fix)-gix)\ix a va cong thiic can a trung gian de pha bd dau gia tri tuyet ddi va cac tim giao diem ciia hai dd thi cau a Hudng ddn S = (x + l ) d x -1 Ddp sd e cau b Hudng ddn S = - j In xdx + j l n xdx Ddpsd 2(1 ) e Bai 14 Hudng ddn Six dung cac phuang phap tinh thi tich khd'i trdn xoay Giao dilm ciia hai dd thi la nghiem ciia he phuang trinh: y = 2x' y = x^ x = 0,y = x = , j ' = V = %\ (2x^)^ - ( x ^ ) ^ dx = 'TC\(AX'^ - x ^ l d x = Tt — x -')• 0^ x Bai 15 Hudng ddn Six dung cac phep toan vl sd phiic cau a Hudng ddn x r., - 22 _ (2 - + (4 + li) + 2i cau b Ddp so X = -— - —i 5 c cau c Hudng ddn Phuong trinh da cho cd A' = - 13 = 12i' Ddp sd X = ± 2V3i 148 _ 256Tt 35 cau d Huang ddn Dat t = x hai nghiem la ^ = - , ^ = Ddp sd X = + Vs , X = ±V2i ta cd phuang trinh bac hai : ^ - i - = vdi Bai 16 Hudng ddn Six dung cac tinh chat ciia so phiic, md dun ciia so phiic cau a Hudng ddn \z\[...]... 26 26 53 5 + — i; 26 26 Trd Idi (a) (c) (b) — + — i ; 26 26 M^ (d) 53 26 5 1 26 , 10-3i Cdu 7 Ket qua 5+i , 4 7 (a) 26 , (c) 25 1; 26 47 25 + — i; 26 26 Trd Idi (a) Cdu 8 Ket qua /u^ 47 2 5 (b) — + — i; 26 26 M^ (d) 47 26 25 1 26 10 + 3i 5+i , - 53 (a) 26 X (c) 53 5 1; 26 5 + — i; 26 26 Trd Idi (b) (b) — + — i ; 26 26 (A^ (d) 53 26 5 1 26 111 HOATD0NG5 HCIGFNG DflN Bfll TflP SGK c + di _ ac... 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... Hudng ddn (2 + 30^ = 4 + 12i + (30^ - - 5 + 12i cau b Hudng ddn Ta cd (2 + 30^ = 8 + 3.4.3i + 3 .2( 30^ + (30^ = 8 + 36/ - 54 - 27 f = - 4 6 + 9 j 1 02 HOAT DONG 6 Bfll TflP BO SUNG Bai 1 Chiing minh 1 2- 3i Bai 2 Chiing minh 1 _2- 3i 2 + 3i~ 13 Bai 3 Chiing minh 2 + 3i 13 1 a-bi « + 6^ a^+b^ Bai 4 Chiing minh z = a + bi ia, b e R) thi phdn thuc cua z la a = -iz + z), phdn do cua z la 6 = — iz - z) 2/ Bai 5... 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... 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... I Cho sd phiic z = 3 + 2i n (a) z = 3-2i D D (b) z + z = 2 (c) z.z = 13 3 2 (d) nghich dao cua z la i n 13 13 Trd Idi (a) (b) (c.) (d) D S D D Cdu 2 Cho sd phiic z = 3 - 2i (a) z = 3 + 2i (b) z + z = 2 D D 109 D (c) z.z = 13 3 (d) nghich dao ciia z la ^ • 13 Trd Idi (a) (b) (c) (d) D D S D 2 i D i • • • 13 Cdu 3 Cho sd phiic z = 3 + 2i (a) z 3" i-li 3 (b) z = 2 + 3i i (c) z 2i 2 3 (d) nghich dao ciia... 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... bi~ a^+b^ cau a Hudng ddn Nhan ca tii va mdu vdi 3 + 2i ^, , 2 +i (2 + 0 ( 3 + 2 0 4 7 Dapso 3 - 2 j = j3 =i3-'l3^cau b Hudng ddn Nhan ca tii va mdu vdi 2 - v3i ^, ,, 2 + V6 2V2-V3 Dapso H 1 7 7 cau c Hudng ddn Nhan ca ttr va mdu vdi 2 + 3i nDapso 15^10 h — 1 13 13 cau d Hudng ddn Nhan ca tur va mdu vdi - i Ddp sd ^ ^ ^ = (5 - 2 0 ( - 0 = - 2 - U Bai 2 Hudng ddn Sir dung tuih chdt ciia sd phiic nghich... 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... 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 ... 26 26 53 + — i; 26 26 Trd Idi (a) (c) (b) — + — i ; 26 26 M^ (d) 53 26 26 , 10-3i Cdu Ket qua 5+i , (a) 26 , (c) 25 1; 26 47 25 + — i; 26 26 Trd Idi (a) Cdu Ket qua /u^ 47 (b) — + — i; 26 ... Cdu Ket qua /u^ 47 (b) — + — i; 26 26 M^ (d) 47 26 25 26 10 + 3i 5+i , - 53 (a) 26 X (c) 53 1; 26 + — i; 26 26 Trd Idi (b) (b) — + — i ; 26 26 (A^ (d) 53 26 26 111 HOATD0NG5 HCIGFNG DflN Bfll... 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 Cho sd phiic z = 12 - v3 i; Izl bang (a) V147 ; (b) V15 (c)^ /21 ; (d) Vl53 Trd Idi (a)