C. BAI LUYEN TAP
22Bai 7: Tim so phuc z ma z
= -1
DS: co 4 so phuc — (1± i) va — (-1± i)
2 2
Bai 8: Cho z = a + bi co cac can bac hai la ±( m+ ni). Tim cac can bac hai ciia -a - bi va a - bi
DS: ± (n - mi) va ± (m - ni)
Bai 9: Giai cac phuong trinh bac hai sau clay trong tap hop cac so phiic C: a) z2 - z + 2 = 0
b) 2z2 - 5z + 4 = 0 (Tot nghiep THPT 2006)
nc , l + iV7 5±i\/7
DS: a) z = b) z = — 2 ; 4 Bai 10: Giai cac phuong trinh :
a)z2 + z+l=0 b)z2-zV3+l=0
nc , -l±iS V3 , 1.
DS: a) z = b )— + —1 2 2 2 Bai 11: Trong C hay giai cac phuong trinh sau day: a)x2-(3-i)x + 4-3i = 0 b)3x2V2 -2x^3+ V2 = 0 DS: a) 2 + i; 1 -2i b)^±i.^
6 6 Bai 12: Giai cac phuong trinh sau:
a) x2 + 3ix + 4 = 0 b) 2x2 - ( 4 + i) x = 1
Bai 13: Giai cac phuong trinh z +— = k trong cac trudng hop sau: a) k = 1 b) k = >/2
DS: a)z=i^ b)z=^(1±l)
Bai 14: Giai cac phuong trinh trong C:
a ) z2+ z = 0 b ) ( z2 + z)2 + 4 ( z2 + z ) - 1 2 = 0
H D : Dat z = x + y i dan den he phucmg trinh hai an x, y
zi = 0; z2 = -1 ; z3 = -1 + V23i -1-7231 „ A+ , n . 1 .73 1 . 7 3 Ket qua: zi = 0; z2 = —1 ; Z 3 =" 2 + 1~ 2_' Z 4 = ~2~1~2~ b) 1,-2, ' 2 2
Bai 15: Lap phucmg trinh bac hai co 2 nghiem: zi = 6-3i va z2 = i DS: z2- ( 6 - 2 i ) z + 6i + 3 = 0
Bai 16: Chung minh rang:
Neu phuong trinh anzn + an.izn'' + ... a2z2 + aiz +ao = 0 voi cac he so thuc co nghiem la zo thi z 0 ciing la nghiem cua phuong trinh.
Bai 17: Giai cac phuong trinh trong tap C:
a) x4 - 3x2 + 4 = 0 b) x4 - 30x2 + 289 = 0
r= •
DS: a ) x = + — + - b ) x = ± 4 ± i 2 2
Bai 18: Giai phuong trinh trong C: xJ + 8 = 0 Bai 19: Cho phucmg trinh 3z4
- 5z3 + 3z2
+ 4z - 2 = 0 a) Chung to rang 1+ i la nghiem cua phuong trinh b) Tim cac nghiem con lai
l + 7 l 3 7 l 3- l DS: b) z2 = 1 - i ; z3 = (adsbygoogle = window.adsbygoogle || []).push({});
b b
Bai 20: Giai phuong trinh z4 + 4 = 0 va bieu dien tap nghiem tren map phang phirc.
Bai 21: Giai cac phuong trinh sau
a) ~r~z =~TT~~ b)((2-i)i + 3 + i)(izA) = 0
1 -1 2 + I 2i
DS: a) — + — i b) - 1 + i, -
25 25 2
Bai 22: Giai cac he phucmg trinh sau voi x, y, z, la so phirc:
f ( 3- i ) x + (4 + 2i)y = 2 + 6i {(2 + i)x + (2 - i)y = 6 {(4 + 2i)x - (2 + 3i)y = 5 + 4i [(3 + 2i)x - (3 - 2i)y = 8 DS: a) x = 1 + i, y = i b) x = 2 + i , y = 2 - i
§ 3 . D A N G L U O N G G I A C A. K I E N T H U t C O B A N hay r = - r coscp = coscp' sincp = - sincp1 r = - r '
- Cho so phuc: z = a + bi voi a, b e R, z * 0, ta co ?
r(coscp + isincp) voi r > 0 la dang luong giac cua /~1 TT a . b <=>r=Va + b , coscp = — . smcp = —
r r hL „M(Z)
cp la mot acgumen cua z vdi sd do radian.
Goc luong giac (Ox, OM) = cp + k2rr
tuc la cac acgumen sai khac k27i vdi k e Z.
- Hai sd phuc bang nhau : Cho hai sd phuc khac 0 la
z = r (coscp + isincp) va z' = r'(coscp' + isincp') (r,r',cp, cp' e R) thi cd:
r = r
coscp = coscp'
sin cp = sin cp'
hay
[cp = cp'+ k2rt .cp = cp'+ (2k + l)rt
- Nhan, chia sd phuc dang luong giac
Neu z = r (coscp + isincp), z' = r'(coscp' + isincp') thi cd:
zz' = rr'[cos(cp + 9') + isin(cp + cp')]
— =— [cos(cp - cp') + isin(cp - 9')], z' * 0
z' r '
Bac biet: z2 = r2 = (cos2cp + isin2cp)
z = r(coscp - isincp) = r (cos(-cp) + isin(-cp)) (adsbygoogle = window.adsbygoogle || []).push({});
- = - (cos(~9) + isin(-cp)) = - (coscp - isincp)
z r r
Cong thuc Moa-vro :
Vdi n la so nguyen, n> 1 thi [r(coscp + isincp)]" = rn(cosncp + isinncp).
Bac biet: (COS9 + isin9)n = cosmp + isinn9