Bài tập hàm biến phức ppt

Baứi taọp haứm bieỏn phửực Trang 1 BI TP CHNG I 1.1. Thc hin cỏc phộp tớnh 1. (5 6 ) (2 4 ) i i + + 2. (2 3 )(4 ) i i + 3. 2 3 (1 ) (1 ) i i + 4. 2 3 5 4 i i + + 5. (3 2 )(1 2 ) 4 3 i i i + 6. (1 )(1 2 ) (2 )(4 3 ) i i i i + + 7. 3 2 4 5 (2 ) i i i + + 8. 3 (1 ) (2 )(1 2 ) i i i + + + 9. 5 (1 2 ) i + 10. 25 2 1 2 i i + 11. 9 8 (1 ) ( 2) i i + 12. 3 2 ( 7 3) ( 7 3) i i + 13. 3 4 i + 14. 5 12 i 15. 2 i 16. 1 2 2 i 1.2. Vit s phc di dng lng giỏc v dng m 1. 4 2. 3 i 3. 2 i 4. 2 2 i + 5. 6 2 i 6. 2 3 2 i + 7. 2 5 i + 8. 12 5 i 1.3. Vit di dng m 1. (2 2 )(3 3 3 ) A i i = + 2. (4 4 )( 1 ) B i i = + + 3. 1 i C i = + 4. 2 6 1 3 i D i + = + 1.4. Tỡm modul ca cỏc s phc 1. 2 3 4 i i 2. (1 3)( 3 ) i i + 3. 2 4 (2 3 ) (3 ) i i + 4. 3 2 (3 4 ) (1 3) i i + + 5. 1 2 2 1 1 i i i i + + 6. 1 1 5 1 1 1 2 i i i + + + + 1.5. Gii cỏc phng trỡnh 1. 2 5 6 0 z z i + = 2. 2 4 z z = 3. 2 2 1 3 i z z i + = + 4. | | 3 z z i = + 5. 2 | | 1 6 2 z i z + + = 1.6. Chng minh vi mi s phc 1 2 , , z z z 1. | | | | z z = 2. | | | | z z = 3. 2 | | . z z z = 4. 1 2 1 2 | | | | | | z z z z + + 5. 1 2 1 2 | | | | | | z z z z 1.7. Chng minh rng 1. N u | | 1 z = thỡ 3 2 | 3 | 4 z . 2. Nu | | 2 z = thỡ | 6 8 | 12 z i + + . 1.8. Cho 1 z i w iz + = + . Chng minh rng nu Im( ) 0 z thỡ | | 1 w . Baứi taọp haứm bieỏn phửực Trang 2 1.9. Chng minh rng nu ( ) n u iv x iy + = + thỡ 2 2 2 2 ( ) n u v x y + = + , vi n l s nguyờn. 1.10. Chng minh rng 2 0 1 2 ( ) 0 n n f z a a z a z a z = + + + + = , nu ( ) 0 f z = , vi ( 0, 1, , ) k a k n = . 1.11. Bng cỏch xột tớch ca 1 1 2 i + v 1 1 3 i + , chng minh 1 1 arctan arctan 2 3 4 + = . 1.12. Biu din qua ly tha ca cos , sin x x 1. cos2 , sin2 x x 2. cos 3 , sin 3 x x 3. cos 4 , sin 4 x x 1.13. Tớnh cỏc s phc 1. 3 (1 3) i 2. 4 1 3. 5 ( 3 ) i 4. 4 ( 2 6) i + 5. 5 (2 2 ) i 6. 7 (1 3) i + 7. 1 2 (1 ) i + 8. 6 ( 3 ) i + 9. 4 1 ( 3 ) i 10. 1 3 ( ) i 11. 3 i 12. 5 1 i 13. 10 1 1 2 2 i + 14. 5 1 1 i i + 15. 4 1 3 1 i i + 16. 3 2 ( 1 ) ( 3 ) i i + + 17. 4 3 (1 3) (2 2 ) i i + 18. 10 1 3 1 3 i i + 1.14. Tớnh v vit di dng m 1. 3 2 2 i + 2. 4 3 i + 3. 5 4 3 i + 1.15. Gii phng trỡnh trong 1. 8 16 0 x = 2. 3 1 0 x + = 3. 4 2 1 0 x x + = 1.16. Cho biu thc 1 3 1 i A i + = . 1. Vit biu thc trờn di dng A x iy = + . 2. Vit dng m ca 1 3 i + v 1 i . Suy ra dng lng giỏc ca A , t ú tớnh 7 cos 12 , 7 sin 12 . 1.17. Tớnh ln lt cn bc 2, 3, 4, 5, 6 ca s phc 1 v biu din cỏc giỏ tr ú trờn ng trũn lng giỏc. 1.18. Cỏc giỏ tr ca 1 n l 2 2 cos sin , 0, 1, , 1 k k k w i k n n n = + = . 1. Tớnh t ng 0 1 1 n w w w + + + . 2. Chng minh rng k w v n k w ( 1, , 1 k n = ) l cp s phc liờn hp v nghch o ca nhau. Baứi taọp haứm bieỏn phửực Trang 3 3. Tớnh 1 1 k n k w w + , 1, , 1 k n = . 4. Tớnh 1 1 1 1 k n k w w + , 1, , 1 k n = . 1.19. Xỏc nh cỏc im trong mt phng Oxy biu din s phc z 1. Re( ) Im( ) z z = 2. | | 3 z < 3. | 1 | 1 z i + 4. Re( ) 2 z i = 5. Re( ) Im( ) 1 z z + < 6. | 2 | 4 z i = 7. 0 Re( ) 1 iz < < 8. Im( ) 3 z i 9. | | | 1 | z i z = 10. | | z z = 11. 2 Im( ) 2 z = 12. | | Re( ) z z = 13. arg z z = 14. | 1 | | 1 | 4 z z + + = 15. arg 3 z = 16. arg( 1) 4 z = 17. | | | | 6 z i z i + + < 18. arg 6 4 z < < 1.20. Xỏc nh ng cong C cho bi phng trỡnh 1. 2 2 , 0z t it t = + < < + 2. 3 2 , 1 2 z t it t = < 3. , 0 i z t t t = < < 4. 2 4 ,z t it t = + < < 5. 1 , 1 z t it = < < + 6. 3 3(cos sin ), 2 2 z t i t t = + < < 7. cos 2 sin , 0 z t i t t = + < < 8. 2 2 sin i sin , 2 2 2 t z t t = + < < 9. 2 1 , 1 0 z t i t t = + 10. 2( ), it z t i ie t = + < < BI TP CHNG II 2.1. Tớnh giỏ tr ca hm ti im 0 z 1. ( ) 3 Im( ) f z z z = a) 0 1 z = b) 0 2 z i = c) 0 1 2 z i = 2. 2 ( ) 2 f z z z i = . a) 0 2 z i = b) 0 1 z i = + c) 0 3 2 z i = 3. 2 ( ) | | 2Re( ) f z z iz z = + a) 0 3 4 z i = b) 0 1 2 z i = + 2.2. Tỡm ph n thc v phn o ca cỏc hm 1. ( ) 5 3 4 f z z i = + 2. ( ) 2 3 f z z z i = + 3. 2 ( ) (1 ) f z z i z = 4. ( ) | | 2 Im( ) f z z iz = 5. ( ) 1 z f z z = + 6. 1 ( )f z z z = Baứi taọp haứm bieỏn phửực Trang 4 2.3. Bit i z re = , tỡm phn thc v phn o ca cỏc hm theo , r 1. ( ) f z z = 2. ( ) | | f z z = 3. 4 ( ) f z z = 4. 2 2 ( ) f z x y iy = + 5. 1 ( )f z z z = + 6. 2 ( ) Re( ) f z z = 2.4. Tỡm min giỏ tr ca cỏc hm 1. ( ) f z z = xỏc nh trong na mt phng trờn Im( ) 0 z > . 2. ( ) Im( ) f z z = xỏc nh trong hỡnh trũn úng | | 2 z . 3. ( ) | | f z z = xỏc nh trong hỡnh vuụng 0 Re( ) 1 z , 0 Im( ) 1 z . 2.5. Tỡm min xỏc nh v min giỏ tr ca cỏc hm 1. ( ) z f z z = 2. ( ) z z f z z z + = 2.6. Tỡm nh B ca tp A qua phộp bin hỡnh ( ) w f z = 1. ( ) f z z = ; a) A l ng thng 3 y = b) A l ng thng y x = 2. ( ) (1 ) f z i z = + a) A l ng thng 2 x = b) A l ng thng 2 1 y x = + 3. ( ) 2 f z z = a) A l na mt phng trờn Im( ) 1 z > b) { : 0 Re( ) 1} A z z = < < 4. ( ) f z iz = a) A l ng trũn | 1 | 2 z = b) { : 1 Im( ) 0} A z z = < < 2.7. Tỡm nh qua phộp bin hỡnh 2 w z = 1. ng thng 1 x = 2. ng thng 2 y = 3. tp : 0 arg 4 A z z = 2.8. Tỡm nh qua phộp bin hỡnh 1 w z = 1. ng thng y x = 2. ng thng 1 x = 3. on thng t im 1 i n im 2 2 i 4. ng trũn | | 2 z = 5. ng trũn | 1 | 1 z = 6. tp { } : 1 Re( ) 2 A z z = < < 2.9. Tỡm gii hn ca cỏc hm s 1. 2 2 lim ( ) z i z z 2. 2 1 lim (| | ) z i z iz + 3. 1 lim z i z z z z + + 4. 2 Im( ) lim Re( ) z i z z z + 2.10. Ch ng minh 1. 0 lim z z c c = ( c ) 2. 0 0 lim z z z z = 3. 0 1 lim z z = 4. 1 lim 0 z z = Baứi taọp haứm bieỏn phửực Trang 5 2.11. S dng kt qu bi 2.10, tỡm cỏc gii hn 1. 2 2 lim ( ) z i z z + 2. 5 lim( 2 1) z i z z + 3. 4 1 lim z i z z i + 4. 2 2 1 1 lim 1 z i z z + 5. 2 2 2 lim (1 2 ) z z iz i z + + 6. 1 lim 2 z iz z i + 2.12. Xột xem cỏc hm s sau cú gii hn khi 0 z hay khụng? 1. Re( ) ( ) Im( ) z f z z = 2. ( ) z f z z = 2.13. Chng t rng cỏc hm s sau liờn tc ti 0 z 1. 2 ( ) 3 2 f z z iz i = + ; 0 2 z i = 2. ( ) 3 Re( ) f z z z i = + ; 0 3 2 z i = 3. 3 1 , | | 1 ( ) 1 3, | | 1 z z f z z z = = ; 0 1 z = 2.14. Xột tớnh liờn tc ca cỏc hm s 1. ( ) f z z = 2. 2 ( ) Im( ) f z z = 2.15. Xột tớnh kh vi ca hm ( ) f z v tớnh o hm (nu cú) 1. ( ) f z z = 2. 2 ( ) ( ) f z z = 3. ( ) .Im( ) f z z z = 4. ( ) . f z z z = 5. 2 ( ) Re( ) f z z = 6. 2 | 1| ( ) z f z e = 7. 5 ( ) f z z z = + 8. ( ) | | . f z z z = 2.16. Chng t cỏc hm s sau khụng gii tớch ti bt k im no trong mt phng phc 1. ( ) Re( ) f z z = 2. ( ) f z y ix = + 3. 2 2 ( ) 2 ( ) f z x y i y x = + + 4. ( ) 4 6 3 f z z z = + 2.17. Cỏc hm s sau gii tớch trong min no ca mt phng phc 1. 2 ( ) f z z = 2. ( ) .Re( ) f z z z = 3. 2 ( ) Im( ) f z z = 4. 2 ( ) f z z zz = + 2.18. Chng minh rng 1. Nu f gii tớch trong min D v | ( ) | f z l hng s trong D thỡ f cng l hng s trong D . 2. Nu f gii tớch trong min D v ( ) 0 f z = thỡ f l hng s trong D . 2.19. Cho i z x iy re = + = v ( ) ( , ) ( , ) f z u x y iv x y = + . Chng minh iu kin C - R tng ng vi 1 u v r r = , 1 v u r r = . 2.20. Tỡm hm gi i tớch ( ) , f z u iv = + bit 1. u x y = + 2. 2 v x y = 3. 2 3 u xy x = + 4. y x v e = 5. 3 2 3 v x xy = 6. arctan y u x = 7. 2 2 ln( ) 2 u x y x y = + + 8. cos 2 x v e y x y = + Baứi taọp haứm bieỏn phửực Trang 6 2.21. Tỡm cỏc hng s , , a b c v d cỏc hm sau gii tớch 1. ( ) ( ) f z x ay i bx cy = + + + 2. ( ) 3 ( 3) f z x y i ax by = + + 3. 2 2 2 2 ( ) ( ) f z x axy by i cx dxy y = + + + + + 2.22. Tỡm phn thc v phn o ca cỏc hm s 1. 3 4 ( ) i f z e + = 2. ( ) cos(2 ) f z i = + 3. ( ) sin 2 f z i = 4. ( ) sh(1 4 ) f z i = 2.23. Tớnh sin z , nu ln(3 2 2) 2 z i = + . 2.24. Chng minh 1. sin( ) sin ch cos sh x iy x y i x y + = + 2. cos( ) cos ch sin sh x iy x y i x y + = + 2.25. Gii phng trỡnh 1. sin 2 z = 2. cos 3 z = 3. 1 z e = 4. 1 z e = . BI TP CHNG III 3.1. Tớnh cỏc tớch phõn 1. ( 3 ) C I z i dz = + , C cú phng trỡnh 2 , 4 1, 1 3 x t y t t = = . 2. (2 ) C I z z dz = , C cú phng trỡnh 2 , 2, 0 2 x t y t t = = + . 3. 2 | | C I z dz = , C cú phng trỡnh 2 1 , , 1 2 x t y t t = = < . 4. Re( ) C I z dz = , C l ng trũn | | 1 z = . 5. C I z dz = , C l ng trũn | | 1 z = . 6. 2 1 2 3 ( ) C I dz z i z i = + + + , C l ng trũn | | 1 z i + = . 3.2. Tớnh cỏc tớch phõn 1. 2 3 ( ) C I x iy dz = + , C l on thng a) ni t 1 z = n z i = . b) ni t 1 z i = + n z i = . 2. 1 C z I dz z + = , C l a) n a phi ng trũn n v t z i = n z i = . b) na trỏi ng trũn n v t z i = n z i = . 3. 2 2 ( ) C I x iy dz = , C cú im u 1 z = v im cui 1 z = , trong hai trng hp sau Baứi taọp haứm bieỏn phửực Trang 7 a) C l na di ng trũn n v. b) C l na trờn ng trũn n v. 4. z C I e dz = , vi C l ng gp khỳc ni 0, 2, 1 i + . 5. C I zdz = theo ellip 2 2 1 4 x y + = t im 2 z = n im z i = , chiu ngc kim ng h. 6. 1 C I dz z = theo ng trũn | | 1 z = t im 1 z = n im 1 z = , trong hai trng hp a) na mt phng trờn. b) na mt phng di. 7. Im( ) C I z i dz = , vi biờn C gm ng trũn | | 1 z = t 1 z = n z i = v on thng t z i = n 1 z = . 8. z C I ze dz = , vi C l biờn ca hỡnh vuụng cú cỏc nh l 0 z = , 1, 1 z z i = = + v z i = . 3.3. Tớnh tớch phõn ( ) C I f z dz = , vi C l biờn ca tam giỏc cú cỏc nh l 0, 1 z z = = v 1 z i = + : 1. ( ) Re( ) f z z = 2. 2 ( ) f z z = 3. ( ) 2 1 f z z = 4. 2 ( ) f z z = 3.4. Tớnh tớch phõn 2 ( 2) C I z z dz = + , vi C cú im u z i = , im cui 1 z = , trong cỏc trng hp: 1. C l ng thng 1 x y + = 2. C l ng gp khỳc ni , 1 , 1 i i + 3. C l parabol 2 1 y x = 4. C l ng trũn | | 1 z = (chiu cựng kim ng h) 3.5. Tớnh tớch phõn 2 ( ) C I z dz = vi C l ng ni t im 0 z = n im 1 z i = + trong trng hp 1. C l on thng. 2. C l ng gp khỳc ni 0, 1, 1 i + . 3.6. Tớnh tớch phõn | | C I z dz = , nu C l 1. on thng ni t im z i = n im z i = . 2. na trỏi ng trũn | | 1 z = ni t im z i = n im z i = . 3. na phi ng trũn | | 1 z = ni t im z i = n im z i = . 3.7. Tớnh tớch phõn C I zdz = t 1 z = n z i = , theo mi ng sau 1. dc theo trc Ox n O , ri dc theo trc Oy n i . 2. dc theo ng thng 1 y x = . 3. d c theo ng thng ng n (1 ) i + ri dc theo ng ngang n i . 3.8. Tớnh tớch phõn | | C I z dz = , vi C l ng trũn | | 1 z i = . Baứi taọp haứm bieỏn phửực Trang 8 3.9. Tớnh cỏc tớch phõn 1. 3 2 0 i z dz + 2. /2 i z i e dz 3. 2 sin 2 i z dz + 4. 0 i z ze dz 5. 1 0 sin i z zdz + 6. 1 2 1 ( 1) i dz z + 3.10. Tớnh cỏc tớch phõn 1. cos C I zdz = , trong ú 2 {( , ) : , 0 1} C x y x y y = = 2. 2 1 C I dz z = , C l elip 2 2 1 ( 1) ( 2) 1 4 x y + + = . 3. 2 3 2 2 C z I dz z z = + + , : | | 1 C z = 4. 2 1 (1 ) C I z dz = + , C l elip 2 2 4 1 x y + = . 3.11. Tớnh cỏc tớch phõn 1. 2 2 C z I dz z i = a) : | | 3 C z = b) : | | 1 C z = 2. 2 1 9 C I dz z = + a) : | 2 | 2 C z i = b) 1 : | 2 | 2 C z i + = 3. 2 2 3 4 C z z i I dz z z + = + a) : | | 2 C z = b) 3 : | 5 | 2 C z + = 4. 2 2 ( 1 ) C z I dz z z i + = a) : | | 1 C z = b) : | 1 | 1 C z i = 5. 3 1 ( 4) C I dz z z = a) : | | 1 C z = b) : | 2 | 1 C z = 6. 3 ( 1) z C e I dz z z = a) : | 2 | 1 C z + = b) 1 : | | 2 C z = c) 1 : | 1 | 2 C z = d) : | | 2 C z = 3.12. Tớnh cỏc tớch phõn 1. 2 2 cos( ) C iz I dz z = + , : | | 4 C z i = 2. 2 sin( / 2) 1 C z I dz z = , : | | 3 C z i+ = 3. 2 2 , C : | | 2 ( 1) z C e I dz z i z z = = + 4. 2 2 , C : | | 2 ( 1)( 3) C z I dz z z z = = + + 5. 3 2 1 , C : | 2 | 4 ( 1) C I dz z z z = = 6. 5 cos2 , C : | | 1 C z I dz z z = = Baứi taọp haứm bieỏn phửực Trang 9 3.13. Tớnh cỏc tớch phõn 1. 2 2 2 sin 4 , : 2 1 C z I dz C x y x z = + = . 2. 2 cos 1 C z I dz z = , C l biờn ca tam giỏc cú cỏc nh 0 z = , 2 2 z i = v 2 2 z i = + . 3. 2 2 4 C z I dz z = + , C l biờn ca hỡnh vuụng cú cỏc nh l 2 z = , 2 z = , 2 4 z i = + v 2 4 z i = + . 4. 2 2 4 iz C e I dz z = , C cú phng trỡnh 2 , 0 2 it z i e t = + . 3.14. Cho 0 t > v C l ng cong trn, kớn, bao quanh im 1 z = . Chng minh rng: 2 3 1 2 2 ( 1) zt t C ze t dz t e i z = + . 3.15. Cho C l na trờn ng trũn | | z R = t z R = n z R = . Chng minh rng: 2 2 2 2 ( 0, 0) imz C e R dz m R a z a R a > > > + . BI TP CHNG IV 4.1. Xỏc nh xem cỏc dóy sau, dóy no hi t, dóy no phõn k 1. 2 1 in in + 2. (1 ) 1 n n i n + + 3. n n i n + 4. 1 2 arctan n e i n + 4.2. Chng t cỏc dóy sau hi t bng cỏch s dng nh lý 4.1 1. 4 3 2 n i n n i + + 2. 1 4 n i + 4.3. Tỡm tng ca cỏc chui sau (nu chui hi t) 1. 1 ( 1) k i k k = + 2. 1 (1 ) k k i = 3. 1 3 2 1 3 k k i = + 4. 1 1 1 4 2 k k i = 5. 1 1 (1 ) k k k i i = + 6. 1 1 1 2 1 2 k k i k i = + + + 4.4. Tỡm bỏn kớnh v hỡnh trũn hi t ca cỏc chui 1. 2 1 ( ) n n z i n = 2. 1 0 ( 2 ) (1 ) n n n z i i + = 3. 1 ( 1) ( 1 ) 2 n n n n z i n = + 4. 2 1 (2 )! ( 1) ( !) n n n z n = 5. 1 n n n z n = 6. 1 1 2 ( 1) ( 2 ) 2 n n n n i z i = + + Baứi taọp haứm bieỏn phửực Trang 10 4.5. Khai trin Taylor cỏc hm s sau trong lõn cn im a v cho bit min khai trin c 1. 1 ( ) 1 f z z = , 3 a i = 2. ( ) z f z e = , a i = 3. 1 ( )f z z = , 1 a = 4. ( ) cos f z z = , / 4 a = 5. 1 ( ) 2 f z z = + , 1 a = , a i = 6. 1 ( ) 1 z f z z = + , 0 a = , 1 a = 4.6. Khai trin Taylor cỏc hm s sau trong lõn cn im a v tỡm bỏn kớnh hi t 1. ( ) ( )( 2 ) i f z z i z i = , 0 a = 2. 2 ( ) 2 3 z f z z z = , 0 a = , 2 a = 4.7. Khai trin Laurent cỏc hm s trong min cho trc 1. cos ( ) , | | 0 z f z z z = > 2. 5 sin ( ) , | | 0 z z f z z z = > 3. 2 1 ( ) , | | 0 z f z e z = > 4. 2 2 ( ) , | 1 | 0 ( 1) z e f z z z = > 5. 2 ( ) sin , | | 0 f z z z z = > 6. 1 2 ( ) , | | 0 z f z z e z = > 7. 2 1 ( ) , 0 | 1 | 1 ( 1) f z z z z = < + < + 8. 2 2 1 ( ) , 0 | | 2 ( 1) f z z i z = < + < + 4.8. Khai trin Laurent hm s 2 1 ( ) ( 1) f z z z = trong cỏc min 1. 0 | | 1 z < < 2. | | 1 z > 4.9. Khai trin Laurent hm s 1 ( ) ( 3) f z z z = trong cỏc min 1. 0 | | 3 z < < 2. | | 3 z > 3. 0 | 3 | 3 z < < 4. | 3 | 3 z > 5. 1 | 4 | 4 z < < 6. 1 | 1 | 4 z < + < 4.10. Khai trin Laurent hm s ( ) ( 1)( 2) z f z z z = + trong cỏc min 1. 1 | | 2 z < < 2. | 1 | 2 z + > 3. 0 | 1 | 3 z < + < 4. 0 | 2 | 3 z < < 4.11. Tỡm v phõn loi cỏc im bt thng cụ lp ca cỏc hm s 1. 3 2 ( ) ( 1) ( 1) z f z z z z + = + 2. 3 1 cos ( ) z f z z = 3. 2 2 2 ( ) ( 1) f z z = + 4. 1 ( ) z f z ze = 5. 5 2 ( ) ( 1) z f z z = 6. 1 ( ) cosf z z i = + 7. 2 3 1 ( ) 2 5 z f z z z = + + 8. 3 1 ( ) ( 1)( 1) z f z z z = + + 9. 2 3 sin ( ) z f z z z = 10. 6 cos cos2 ( ) z z f z z =