.K IEN THLTC CO BN §1 SOPH UC

C. BAI LUYEN TAP

A .K IEN THLTC CO BN §1 SOPH UC

A - Tap hop so phuc: C

- So phuc (dang dai so): z = a + bi (a, b e R, i la don v i ao, i2 - - 1 ) .

Goi a la phan thuc, b la phan ao cua z. z la sd thuc o> phan ao cua z bing 0 o z = a z la sd ao o phan thuc ciia z bing 0 <=> z = bi z = 0 la sd phuc duy nhat vua la so thuc vua la sd ao.

- Bieu dien hinh hoc: Sd phuc z = a + bi, (a, b e R) duoc bieu dien bdi diem M(a;b) hay bdi vecto u (a;b) trong mat phing toa do Oxy gpi la mat phang phuc. True thuc la true hoanh va true ao la true tung

- Hai sd phuc bang nhau a + bi = a' + b'i (a, b, a', b' e R) o a = a', b = b'

- Cong, tru sd phuc, nhan hai sd phuc: (a + bi) + (a' + b'i) = (a + a') + (b + b')i (a + bi) - (a' + b'i) = (a - a') + (b - b')i (a + bi) (a' + b'i) = (aa'-bb') + (ab' + ba')i (a, b, a', b' e R)

- Sd phuc lien hiep cua sd phuc: z = a + bi, (a, b e R) la z = a - bi Ket qua z = z;z + z' = z + z';zz' = z.z1

z la sd thuc <=> z = z ; z la sd ao <=> z = - z - Mdn dun cua sd phuc: z = a + bi (a, b e R)

| z | = v V + b2 = V Z I = | 0 M |

o

rM(a+bi)

->

Ket qua | z| > 0 vdi moi z e C va I z| = 0 <=>„z = 0

lzz'| = | z | . | z ' | ; | z |2= | z2|

Iz + z'l < | z | + |z'| vdi moi z, z ' e C

- Sd phuc nghich dao cua z (z * 0): z- 1 = - = z

Chia hai so phuc: Phep chia cua z' cho z * 0: — = z'.z"1 = — =7Lt-

Cu the, cho hai sd phuc z = a + bi va z' = a' + b'i * 0 thi cd: z z' aa'+ bb' a' b - ab' . • + —^ — l a ^{'2+ b '2 L'2+b'2} -BDHSG DSGT12/2- ₁₃₉

Ket qua: Voi z * 0, — = w <=> z' = wz; '

z \z J z

B. P H A N D A N G T O A N

DANG 1. PHEP TOAN VE s 6 PHUC Nguyen tac chung de tinh toan:

- Do co cac tinh chat giao hoan, phdi hop nen quy tac cong, tru so phuc la cong rieng, trir rieng cac phan thuc va phan ao.

- Do co cac tinh chat giao hoan, ket hop ciia phep nhan, tinh chat phan phoi ciia phep nhan doi voi phep cong nen phep nhan hai so phuc duoc thuc hien theo quv tac nhan da thtic thdng thudng rdi thay:

i2 = - 1 . i3 = i2 i = - i , i4 = i2 . i2 = 1,..., i4"1 = 1. i4 m + 1 = i, i4 m + 2 = - 1 , i4 m + 3 = - 1. - De tinh —, ta nhan tir va mau cho sd phiic lien hiep ciia mau z' - Tdng quat la tinh gon tirng budc, dung cac gia tri cua in va ha bac cac

luy thira, phan tich thuan led, tinh gon mau neu cd trudc khi nhan sd phiic lien hiep,... Chu y: - I z 12 * z2 ; I z 12 = z2 <=> z la sd thuc

- De kiem tra sd phiic la sd thuc hay sd ao, ngoai each tinh ra cu the phan thuc, phan ao thi ta cd the tim sd phiic lien hiep ciia nd rdi so sanh. - Cd the dirng cdng thiic tinh tdng cac cap sd nhan:

1 ^

Ui + U2 + U3 + ... + un = U). —, q * 1, hoac hang dang thiic: l - q

an- bn = ( a- b ) ( an _ 1 + an _ 2. b + an _ 3. b2 + ... + a bn - 2 + bn _ 1) Vi du 1: Cho z = (2a - 1) + (3b + 5)i vdi a, b e R

Tim cac so a, b de: a) z la sd thuc b) z la sd ao Giai 5 a) z la so thuc <=>3b + 5 = 0<=>b = — 3 b) z la sd ao o 2a - 1 = 0 <=> a = — 2

Vi du 2: Tim z va tinh I z| vdi: a) z = - 2 + i V3 b) z = V2 - 2i Giai

a) So phiic lien hiep z = - 2 - i S mddun I z| = Va2 + b2 = ^4 + 3 = ^7 b) So phiic lien hiep z = V2 + 2i mddun | z | = V2 + 4 = V 6 = z + z' 140 -BDHSG DSGT12/2-

V j du 3: Tim so phuc z thoa man tung truong hop: a) | z | = 2 va z la so ao

b) I z | = 5 va phan thuc cua z bang hai lan phan ao ciia no. Giai

Gia sir z = a + bi, voi a, b e R

. , J z = 2 L /a 2+ b2 = 2 [a = 0 a) Ta co: 1 <=> <^ v <=> 1 a = 0 [a = 0 [b = ±2 Vay: z = ±2i MT , JN = 5 JVa2+b2 =5 fa = 2b }a = -2V5. Ja = 2S b) Ta co: <1 o <^ <=> <^ <=> < hay <^ [a = 2b [a = 2b [b = ±V5 [b = -V5 [b = V5 Vay co hai so phiic can tim: z = - 2 V5~ - i V5 , z = 2 V5 + i %/5

V i du 4: Chung minh rang hai so phiic phan biet z\, z2 thoa man dieu kien I I I I Z + Zo '

I zi I = |z21 khi va chi khi — la sd ao.

Zl _ Z2 Giai Vdi dieu kien z\ * z2, ta cd: Zl + Z2 la sd ao Zl + Zz

+ zi + z zi + z

Z-^ Z2 Z^ Z2 \^ Z-^ i. o ( zx+ z2) ( z1- z2) + (z1 - z2) ( z , + z2) = 0

<=> 2(Zj Zj - z2 z2) = 0 <=> |z, | = |z21 V i du 5: Tim nghich dao ciia sd phiic sau:

a) z = 3 + 4i b) z = -3 - 2i Giai Ta cd: z. z = I z 12 vdi z = a + bi thi: z. z = a2 + b2 - z a2+ b2 k . 1 z 3 - 4 i 3 4 . Ap dung: a) - = 5- = = 1 z a2+ b2 25 25 25 1_ z -3 + 2i _ 3 _2_. z ~ a2 + b2 ~ 13 " 13 + 1 31 V i du 6: Thuc hien cac phep tinh sau:

A = I B = ^ i C = ^ i ( l + i ) ( 4 - 3 i ) 4 + 3i 8 - 6 i Giai 1 _ ( l - i ) ( 4 + 3i) _ 7 1 . ( l + i ) ( 4 - 3 i ) ( l + T).(16 + 9 ) ~ 5 0 5 01 -BDHSG DSGT12/2- 141

B = z5 + 6 i _ ( - 5 + 6 i ) ( 4- 3 i ) 4 + 3i 25 2 39. — + — i 25 25 r= 7- 2 i=( 7 + 2i)(8 + 6 i ) _ l l _ 29. 8- 6 i 100 _ 25 + 50 1

V i du 7: Thuc hien phep tinh:

1^ 2i (•7 1 ) _{B =} ^{f l + i Y} {' - 7 ) ^{B =} f l + i Y {' - 7 ) _{U - i J} 33 + ( l - i )1 0 + (2 + 3 i ) ( 2 - 3 i ) + C = 1 + (1 + i) + (1 + i )2 + (1 + i )3 + + ( i + i)20 Giai a) T a c 6 : i7 = i6. i = ( i2)3. i = ( - l )3. i = - i Nen A = — I r - 1 2i _i7) 1_ 2i ^{- i +} 1 1 "2 + 2 i2 2 2 ^{= - 1} b)Taco: i ± A = £ ± j £ = 1 + *? + 2 l = I z l l g i = j l - i 1 - i2 Nen: ^{1 + i} l - i \33 ^{1 + 1} ;2\16 = i3 3 = ( i2)1 6. i = i . Va (1 - i )2 = 1 + i2 - 2i = - 2 i

Nen (1 - i )1 0 = ( - 2 i )5 = -32i. Tu do tinh dugc B = 13 - 32i

c) Dung cdng thuc tdng cua cap sd nhan co 21 sd hang:

C = u: l - q ²¹ 1. ^{l - ( l + l )2 1 l - ( l + i )2 1} l - q l - ( l + i) Ta cd: (1 + i )2 = 1 + i2 + 2i = 2i Nen: (1 + i )2' = (1 + i). (1 + i )2 0 = ( i + i } ( 2 i )' o = - ( l + i ) . 2, 0 = - 21 0 - i . 2 '0 Do do: C = 1 + ( 2 l ° _+ i 2 l 0 ) = - 21 0 + (21 0 +1).! Vay C = -210 + (210 + l).i

V i du 8 : H o i m 6 i so sau day la sd thuc hay sd ao (z la sd phuc tuy y cho trudc sao cho bieu thuc xac dinh)

z - z a" a) z^ + ( z )2

b)

Ta tinh cac sd phiic lien hiep:

a) z2 + (z)2 = z2 + z2 = z2 + (z)2 Vay z2 + (z)2 la so thuc. z - z z3 + ( i )3 Giai b) ^{z - z z - z} z3 + (z)3 ( z )3+ z3 z3+ ( z )3 142 _{-BDHSG DSGT12/2-}

z — z

Vay — =—• la so ao. z3

+ ( z )3

(z - z')2 = (z - z').(z - z')

V i du 9: a) Chung minh ring so du cua da thuc Pn(z) chia cho z - z0 bang Pn(Zo)

b) Phep chia da thuc Pn(z) cho z - i duoc sd du la i, chia cho z + i duoc so du la 1+ i. Tim sd du cua phep chia Pn(z) cho z2

+ 1. Giai

a) Gia su phep chia Pn(z) cho z - z0 duoc thuong Qn-i(z) va sd du r e C : P„(z) = (z - z0).Qn_,(z) + r. D l thay r = Pn(z0)

b) So du trong phep chia P„(z) cho z2 + i phai co dang mz + n vdi m. n e C. Theo a) ta cd : Pn(i) = i va Pn(-i) = 1 + i.

,, , , , f m i + n = i _ .,. , A . i 1 Do do ta co he: < Giai he duoc m = —, n = — + l

[ - m i + n = 1 + i 2 2 Vay sd du: —. z + — + i

2 2

DANG 2. BIEU DIEN VATAP HOP DIEM Bieu dien so phuc:

- Neu z bieu dien bdi u va z' bieu dien bdi u ' thi z + z' bieu dien bdi u + u', va z - z' bieu dien bdi u - u '

Neu z. z'bieu dien bdi M , M ' thi z + z' duoc bieu dien bdi OM + O M ' z - z ' duoc bieu dien bdi O M - O M ' = M ' M

Neu k la sd thuc, z bieu dien bdi u thi kz bieu dien bdi k u .

Neu k la sd thuc, z bieu dien bdi diem M thi kz bieu dien bdi k O M . Tap diem bieu dien so phirc :

Cho z = x + yi: M(x, y) hay u (x,y) z' = x' + y'i: M'(x', y') hay u '(x', y') thi cd :

I zl = R. R > 0 <=> x2 + y2 = R2 hay O M = R I z| < R, R > 0 o x2 + y2 < R2 hay O M < R | z - ( a + b i ) | = R, ( R > 0 ) <=> (x - a)2 + (y - b)2 = R2

hay I M = R vdi I(a, b) I z - (a + bi) I < R. (R > 0)

<=> (x - a)2 + (y - b)2 < R2 hay I M < R vdi I(a, b)

| w - (x + yi) | = | w - (x' + y'i) I, N bieu diin w <=> N M = N M ' . - Cac loai phuong trinh, dieu kien djnh nghia:

Ax + By + C = 0, A2 + B2 * 0: dudng thing y = ax2 + bx + c : parabol

(x - a)2 + (y - b )2 = R2: ducmg trdn tam I(a,b), ban kinh R (x - a)2 + (y - b )2 < R2: hinh trdn tam I(a. b), ban kinh R

M F i + M F2 = 2a, F i F2 = 2 c< 2 a : elip

I M F i - MF21 = 2a, F i F2 = 2c > 2a : hypebol

M I = MJ: trung true cua doan IJ

- H i n h dang tap diem:

Vdi tam giac thi cd the deu, can, vudng, vudng can

Vdi tu giac thi cd the hinh vudng, hinh thoi,..., tii giac ndi tiep,... Viec dinh dang phai dua vao dd dai hay gdc dac biet.

V i du 1: Cho cac sd phuc: 2 + 3i; 1 + 2i, 2 - i

a) Bieu dien cac sd dd trong mat phang phuc

b) Viet sd phiic lien hiep cua moi sd dd va bieu dien chung trong mat

phang phiic. Viet sd ddi cua mdi sd phiic do va bieu dien chung trong mat phang phiic.

Giai a) Cac diem A(2;3) bieu dien so phiic 2 + 3i,

B(l;2) bieu dien so phiic 1 + 2i, C( 2 ; - l ) bieu dien so phiic 2 - i

b) Cac sd phiic lien hop cua 2 + 3i, 1 + 2i, 2 - i

lan luot la 2 - 3i, 1 - 2i, 2 + i duoc bieu dien

bdi cac diem: A(2;-3), B( l; - 2 ) , C(2;l)

Cac sd ddi cua 2 + 3i, 1 + 2i, 2 - i lan luot la:

- 2 - 3i, - 1 - 2i, - 2 + i duoc bieu dien bdi

cac diem:A'(-2;- 3), B ' ( - l ; - 2), C'(-2;l)

V i du 2: Tren mat phang phiic tim tap hop diem bieu dien cac sd phiic z thoa

/ 3 j B TT — — [ I i .r A: .A

man dieu kien: a) I z I = 1 b) |z| < 2 Giai

Dat: z = x

phiic. Theo gia thiet:

yi va M ( x ; y) la diem bieu dien sd phiic z tren mat phang = 1 o i> / x2+ y2 l « x2 + y2 1

Vay tap hop cac diem M la dudng trdn tam O ban kinh R = 1. b) Theo gia thiet: I z| < 2 o ^ x2 + y2 < 2 o x2 + y2 < 4

Vay tap hop cac diem M la hinh tron tam 0 , ban kinh R = 2

1 o h "x

Vi du 3: Xac dinh tap hop cac diem trong mat phang phirc bieu dien cac sd phuc z thoa man tirng dieu kien sau:

a) I z - i | 1 _b) 1

|z +1 Giai

a) Cach 1: Goi I la diem bieu dien sd i , M la diem bieu dien so z

Ta cd: _{z - i} I M = I M = M I . Do do: z - i I = 1 <=> M I = 1

Vay quy tich cua M la dudng trdn tam 1(0,1) cd ban kinh bang 1. Cach 2: Gia sir: z = x + y i

Taed: | z - i | = 1 <=> | x + ( y - l ) i |2

= 1 o x2

+ ( y - 1 )2 = 1

Vay tap hop cac diem bieu dien sd phirc z la dudng trdn tam 1(0; 1), ban k i n h R = 1.

b) Cach 1: Goi M la diem bieu dien so z, I la diem bieu dien sd i , J la diem bieu dien so - i thi 1(0, 1) va J(0, -1)

Ta cd: ^{I M}

J M ^{Do dd:} _{z +1} ^{1 o I M = J M o M I = MJ o M} nam tren dudng trung true cua IJ o M thudc true thuc Ox.

Cach 2: Neu z = x + y i (x, y e R), ta cd:

1 z + i|<=> | x + ( y - l ) i | = | x + ( y+ l) i |

o y 0 o z la sd thuc y A

->

Vi du 4: Xac dinh tap hop diem bieu dien sd phirc z thoa dieu kien:

o

a) z la sd ao b) I zl = I z Giai

3 + 4i

a) Gia sir: z = x + y i thi: z2 = (x + y i )2 = x2 - y2 + 2xyi Ta co z2 l a s 6 a o o x2- y2 = 0 o( x - y ) (x + y) = 0

<=> x - y = 0 hay x + y = 0. Vay tap hop cac diem bieu dien so phuc z la hai dudng phan giac x ± y = 0.

b) 3 + 4 i = z - 3 + 4i I z - 3 - 4i | nen neu goi A la diem bieu dien so 3 + 4i, M la diem bieu dien sd z thi I z - 3 + 4i I A M

Do do I z | z - 3 + 4 i | « . OM = A M

nen M nam tren dudng trung true cua doan thing OA vdi A(3;4) V i du 5: Tim tap hop cac diem bieu dien sd phuc z thoa man dieu kien:

a) 21 z - i | 2 i | _b) Giai: ( z z )2 a) Goi z = x + yi, x, y e R. Ta cd: 21 z - i I = I z - z + 2i I o 2| x + ( y- l) i | = 2| ( y + l) i | » x2 + (y - l )2 = (y + l )2 X2 ! X2

<=> y = — Vay tap hop can tim la parabol y = —

4 4 b) Goi z = x + yi, x, y e R. Ta cd: I z2 - (z )21 4 <=> 14xyi I = 4

o | x y |

Vay tap hop can tim la hai hypebol y

1 <=> xy = 1 hoac xy = - 1

1 . 1 — va y = — — va y = —

x x

V i du 6: a) Trong mat phang phuc. cho ba diem A, B, C khdng thang hang theo thu tu bieu dien cac sd phuc Z\, z2, Z3. Hdi trong tam cua tam giac ABC bieu dien sd phuc nao?

b) Xet ba diem A, B, C cua mat phang phuc theo thu tu bieu dien ba so

Z2I Z3I

phuc phan biet z\, z2, Z3 thoa man I z\ I

Chung minh rang A, B, C la ba dinh cua mdt tam giac deu khi va chi khi:

z, + z2 + z3 = 0

Giai

a) G la trong tam cua tam giac ABC khi va chi khi:

OG = - (OA + OB + OC)

V i OA,OB,OC theo thu tu bieu dien Z\, z2, z3 nen G bieu dien so phuc - (z, + z2 + z3)

b) Ba diem A, B, C thudc mdt dudng trdn tam O nen tam giac ABC la tam

giac deu khi va chi khi trong tam G cua nd triing vdi tam dudng trdn ngoai tiep, tire G = 0 hay z\ + z2 + z3 = 0

V i du 7: Goi M , M ' theo thir tu la cac diem cua mat phang phuc bieu dien so 1 + i

z # 0 va z' la gdc toa do)

z . Chirng minh rang tam giac O M M ' la \udng can (0

Giai: _yA Taed: OM =|z| O M ' ^{1 + i} M M = O M ' - O M 2 1 + i zl - • _{2 '} V2, 2 ^M(z)

Do I z | * 0, suy ra tam giac O M M ' la tam giac vuong can dinh M ' .

Vi du 8: Cac vecto u , u ' trong mat phang phuc theo thir tu bieu dien cac so phuc z, z'. Chirng minh:

a) Tich vd hudng u . u' thoa man: u . u' = — (zz '+zz')

- - - - z' •

b) Neu u * 0 thi u , u ' vudng gdc khi va chi khi — la sd ao. z

c) u , u ' vudng gdc khi va chi khi I z + z' I = I z - z' I Giai

a) Viet z = x + yi, z' = x' + y'i (x, y, x', y' e R) thi: u . u ' = xx' + yy'

va: zz' + zz' = (x - yi)(x' + y'i) + (x + yi) (x' - y'i) = 2(xx' + yy')

Nen: u.u' = — (zz' + zz') 2 b) u . u ' = 0 o zz' + z z ' = 0. Do dd: z' 2 u . u ' = 0 <=> — + •= z z 0 o — + — = 0 <=> — la sd ao. c) | z + z' I z - z" | » (z + z') (z + z') = (z - z') (z - z') <=>zz' + zz' = 0 « u , u' vudng gdc. Vay u , u ' vudng gdc khi va chi khi I z + z' I = I z - z' I

Vi du 9: A, B, C, D la bdn diem trong mat phang phuc bieu dien theo thu tu cac so: - 1 + i, - 1 - i, 2i, 2 — 2i

a) Tim cac sd z\, z2, Z3, Z4 theo thu tu bieu dien bdi cac vecto ATj,AD,BC,BD

z z

b) Tinh — , — va tir dd suy ra A, B, C, D ciing nam tren mdt dudng trdn. z2 z4

Tam ducmg trdn dd bieu dien sd phuc nao? Giai AC bieu dien sd phuc z\ = 1 + i AD la bieu dien sd phuc z2 = 3 - 3i

z2 3 - 3 i 3 Do do: AC vuong goc A D

BC bieu dien sd phuc Z3 = 1 + 3i BD bieu dien sd phuc Z4 = 3 - i

V i ^ ^{l + 3i} i la sd ao nen B C . BD '