C\u00e2u 7:<\/b>\u00a0Cho z = -1 + 3i . S\u00f4\u0301 ph\u01b0\u0301c w =\u00a0iz\u2212<\/i>\u00a0+ 2z b\u0103\u0300ng<\/p>\n
A. 1 + 5i\u00a0\u00a0\u00a0B. 1 + 7i\u00a0\u00a0\u00a0C. \u2013 1 + 5i \u00a0\u00a0\u00a0D. \u2013 1 + 7i<\/p>\n
C\u00e2u 8:<\/b>\u00a0Cho z = 1 + 2i . Ph\u00e2\u0300n th\u01b0\u0323c va\u0300 ph\u00e2\u0300n a\u0309o cu\u0309a s\u00f4\u0301 ph\u01b0\u0301c w = 2z +\u00a0z\u2212<\/i>\u00a0la\u0300<\/p>\n
A. 3 va\u0300 2 \u00a0\u00a0\u00a0B. 3 va\u0300 2i \u00a0\u00a0\u00a0C. 1 va\u0300 6 \u00a0\u00a0\u00a0D. 1 va\u0300 6i<\/p>\n
C\u00e2u 9:<\/b>\u00a0Cho s\u00f4\u0301 ph\u01b0\u0301c z tho\u0309a ma\u0303n (1 + 2i)z +\u00a0iz\u2212<\/i>\u00a0= 2i . Khi \u0111o\u0301 ti\u0301ch z.iz\u2212<\/i>\u00a0b\u0103\u0300ng<\/p>\n
A. \u2013 2 \u00a0\u00a0\u00a0B. 2 \u00a0\u00a0\u00a0C. \u2013 2i \u00a0\u00a0\u00a0D. 2i.<\/p>\n
C\u00e2u 10:<\/b>\u00a0M\u00f4\u0111un cu\u0309a s\u00f4\u0301 ph\u01b0\u0301c z tho\u0309a ma\u0303n 2z + 3(1 – i)iz\u2212<\/i>\u00a0= 1 – 9i la\u0300<\/p>\n
A. 5 \u00a0\u00a0\u00a0B. 13 \u00a0\u00a0\u00a0 C. \u221a5 \u00a0\u00a0\u00a0D. \u221a13<\/p>\n
C\u00e2u 11:<\/b>\u00a0Cho hai s\u00f4\u0301 ph\u01b0\u0301c z1<\/sub>, z2<\/sub>\u00a0tho\u0309a ma\u0303n |z1<\/sub>| = |z2<\/sub>| = |z1<\/sub>\u00a0+ z2<\/sub>| = 1 . Khi \u0111o\u0301 |z1<\/sub>\u00a0– z2<\/sub>| b\u0103\u0300ng<\/p>\n A. 0 \u00a0\u00a0\u00a0B. 1\u00a0\u00a0\u00a0C. 2\u00a0\u00a0\u00a0D. \u221a3<\/p>\n C\u00e2u 12:<\/b>\u00a0T\u00e2\u0323p h\u01a1\u0323p ca\u0301c \u0111i\u00ea\u0309m bi\u00ea\u0309u di\u00ea\u0303n s\u00f4\u0301 ph\u01b0\u0301c z tho\u0309a ma\u0303n |z + 1 – 2i| = 2 la\u0300<\/p>\n A. \u0110\u01b0\u01a1\u0300ng tro\u0300n t\u00e2m I(1; -2) ba\u0301n ki\u0301nh R = 2<\/p>\n B. \u0110\u01b0\u01a1\u0300ng tro\u0300n t\u00e2m I(1; -2) ba\u0301n ki\u0301nh R = 4<\/p>\n C. \u0110\u01b0\u01a1\u0300ng tro\u0300n t\u00e2m I(-1; 2) ba\u0301n ki\u0301nh R = 2<\/p>\n D. \u0110\u01b0\u01a1\u0300ng tro\u0300n t\u00e2m I(-1; 2) ba\u0301n ki\u0301nh R = 4<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 7:<\/b><\/p>\n Ta c\u00f3: z = -1 + 3i =>\u00a0z\u2212<\/i>\u00a0= -1 – 3i =>\u00a0iz\u2212<\/i>\u00a0= – i – 3i2<\/sup>\u00a0= 3 – i<\/p>\n Suy ra: w = 2z +\u00a0z\u2212<\/i>\u00a0= 3 – i + 2(-1 + 3i) = 1 + 5i<\/p>\n C\u00e2u 8:<\/b><\/p>\n Ta c\u00f3: w = 2z +\u00a0z\u2212<\/i>\u00a0= 2(1 + 2i) + (1 – 2i) = 3 + 2i<\/p>\n V\u1eady ph\u1ea7n th\u1ef1c c\u1ee7a w l\u00e0 3, ph\u1ea7n \u1ea3o c\u1ee7a w l\u00e0 2<\/p>\n C\u00e2u 9:<\/b><\/p>\n \u0110\u1eb7t z = a + bi(a, b \u2208 R). Ta c\u00f3:\u00a0z\u2212<\/i>\u00a0= a – bi v\u00e0 (1 + 2i)z = (1 + 2i)(a + bi) = a + bi + 2ai + bi2<\/sup>\u00a0= a – 2b + (2a + b)i<\/p>\n Do \u0111\u00f3: (1 + 2i)z +\u00a0z\u2212<\/i>\u00a0= 2i <=> a – 2b + (2a + b)i + a – bi = i<\/p>\n <\/p>\n Suy ra z = 1 + i. V\u1eady z.z\u2212<\/i>\u00a0= |z\u2212<\/i>|2<\/sup>\u00a0= 12<\/sup>\u00a0+ 12<\/sup>\u00a0= 2<\/p>\n C\u00e2u 10:<\/b><\/p>\n \u0110\u1eb7t z = a + bi (a, b \u2208 R). Ta c\u00f3:\u00a0z\u2212<\/i>\u00a0= a – bi v\u00e0 (1 – i)z\u2212<\/i>\u00a0= (1 – i)(a – bi) = a – bi – ai + bi2<\/sup>\u00a0= a – b – (a + b)i Do \u0111\u00f3 2z + 3(1 – i)z\u2212<\/i>\u00a0= 1 – 9i <=> 2(a + bi) + 3[a – b – (a + b)i] = 1 – 9i<\/p>\n <=> (5a – 3b) – (3a + b)i = 1 – 9i<\/p>\n <\/p>\n Suy ra z = 2 + 3i. V\u1eady:<\/p>\n <\/p>\n C\u00e2u 11:<\/b><\/p>\n C\u00e1ch 1: \u0110\u1eb7t z1<\/sub>\u00a0= a1<\/sub>\u00a0+ b\u00a01<\/sub>i, z2<\/sub>\u00a0= a2<\/sub>\u00a0+ b2<\/sub>i (a1<\/sub>, a2<\/sub>, b1<\/sub>, b2<\/sub>\u00a0\u2208 R). Ta c\u00f3:<\/p>\n |z1<\/sub>| = |z2<\/sub>| = 1<\/p>\n <\/p>\n |z1<\/sub>| + |z2<\/sub>| = => (a1<\/sub>\u00a0+ a2<\/sub>)2<\/sup>\u00a0+ (b1<\/sub>\u00a0+ b2<\/sub>)2<\/sup>\u00a0= 1 => 2(a1<\/sub>a2<\/sub>\u00a0+ b1<\/sub>b2<\/sub>) = -1<\/p>\n Do \u0111\u00f3:<\/p>\n <\/p>\n C\u00e1ch 2: Ta c\u00f3: |z1<\/sub>| = |z2<\/sub>| = 1 => z1<\/sub>z1<\/sub>\u2212<\/i>\u00a0= z2<\/sub>z2<\/sub>\u2212<\/i>\u00a0= 1<\/p>\n |z1<\/sub>| + |z2<\/sub>| = 1<\/p>\n <\/p>\n Do \u0111\u00f3<\/p>\n <\/p>\n V\u1eady |z1<\/sub>| – |z2<\/sub>| = \u221a3<\/p>\n C\u00e2u 12:<\/b><\/p>\n \u0110\u1eb7t z = a + bi(a, b \u2208 R). Ta c\u00f3: z + 1 – 2i = (a + 1) + (b – 2)i. Do \u0111\u00f3:<\/p>\n |z + 1 – 2i| = 2 <=> (a + 1)2<\/sup>\u00a0+ (b – 2)2<\/sup>\u00a0= 4<\/p>\n V\u1eady t\u1eadp h\u1ee3p \u0111i\u1ec3m M bi\u1ec3u di\u1ec5n s\u1ed1 ph\u1ee9c z l\u00e0 \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m I(-1 ;2), b\u00e1n k\u00ednh R = 2<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 7:\u00a0Cho z = -1 + 3i . S\u00f4\u0301 ph\u01b0\u0301c w =\u00a0iz\u2212\u00a0+ 2z b\u0103\u0300ng A. 1 + 5i\u00a0\u00a0\u00a0B. 1 + 7i\u00a0\u00a0\u00a0C. \u2013 1 + 5i \u00a0\u00a0\u00a0D. \u2013 1 + 7i C\u00e2u 8:\u00a0Cho z = 1 + 2i . Ph\u00e2\u0300n th\u01b0\u0323c va\u0300 ph\u00e2\u0300n a\u0309o cu\u0309a s\u00f4\u0301 ph\u01b0\u0301c w = 2z +\u00a0z\u2212\u00a0la\u0300 A. 3 va\u0300 2 \u00a0\u00a0\u00a0B. […]<\/p>\n","protected":false},"author":3,"featured_media":26742,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[157],"tags":[1447,1446],"yoast_head":"\n\n\n
\n 7-A<\/td>\n 8-A<\/td>\n 9-B<\/td>\n 10-D<\/td>\n 11-D<\/td>\n 12-C<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n