Bài tập toán cao cấp Tập 1 part 2 docx

1.4. Biˆe ’ udiˆe ˜ nsˆo ´ ph´u . cdu . ´o . ida . ng lu . o . . ng gi´ac 27 T`u . d ´othudu . o . . c z 2 =  2(1 + cos ϕ)  cos  − ϕ 2  + i sin  − ϕ 2  v`a do vˆa . y w =  2(1 + cos ϕ)  2(1 + cos ϕ) ×  cos ϕ 2 + i sin ϕ 2  cos  − ϕ 2  + i sin  − ϕ 2  = cos ϕ + i sin ϕ.  V´ı d u . 3. 1) T´ınh ( √ 3+i) 126 2) T´ınh acgumen cu ’ asˆo ´ ph´u . c sau w = z 4 − z 2 nˆe ´ u argz = ϕ v`a |z| =1. Gia ’ i. 1) Ta c´o √ 3+i =2  cos π 6 + i sin π 6  .T`u . d´o ´a p d u . ng cˆong th ´u . c Moivre ta thu du . o . . c: ( √ 3+i) 126 =2 126  cos 126π 6 + i sin 126π 6  =2 126 [cos π + i sin π]=−2 126 . 2) Ta c´o w = z 4 −z 2 = cos 4ϕ + i sin 4ϕ −[cos 2ϕ −i sin 2ϕ] = cos 4ϕ − cos 2ϕ + i(sin 4ϕ + sin 2ϕ) = −2 sin 3ϕ sin ϕ +2i sin 3ϕ cos ϕ = 2 sin 3ϕ[−sin ϕ + i cos ϕ]. (i) Nˆe ´ u sin 3ϕ>0(t´u . c l`a khi 2kπ 3 <ϕ< (2k +1)π 3 , k ∈ Z )th`ı w = 2 sin 3ϕ  cos  π 2 + ϕ  + i sin  π 2 + ϕ  . (ii) Nˆe ´ u sin 3ϕ<0(t´u . c l`a khi (2k − 1)π 3 <ϕ< 2kπ 3 , k ∈ Z )th`ı w =(−2 sin 3ϕ)[sin ϕ −icos ϕ]. 28 Chu . o . ng 1. Sˆo ´ ph´u . c Ta t`ım da . ng lu . o . . ng gi´ac cu ’ a v = sin ϕ − i cos ϕ.Hiˆe ’ n nhiˆen |v| =1. Ta t´ınh argv argv = arctg  −cos ϕ sin ϕ  = arctg(−cotgϕ) = arctg  − tg  π 2 − ϕ  = arctg  tg  ϕ − π 2  = ϕ − π 2 · Nhu . vˆa . ynˆe ´ u sin 3ϕ<0th`ı w =(−2 sin 3ϕ)  cos  ϕ − π 2  + i sin  ϕ − π 2  . (iii) Nˆe ´ u sin 3ϕ =0⇒ ϕ = kπ 3 ⇒ w =0. Nhu . vˆa . y argw =            π 2 + ϕ nˆe ´ u 2kπ 3 <ϕ< (2k +1)π 3 , khˆong x´ac di . nh nˆe ´ u ϕ = kπ 3 , ϕ − π 2 nˆe ´ u (2k − 1)π 3 <ϕ< 2kπ 3 ·  V´ı d u . 4. Ch´u . ng minh r˘a ` ng 1) cos π 9 + cos 3π 9 + cos 5π 9 + cos 7π 9 = 1 2 . 2) cos ϕ + cos(ϕ + α)+cos(ϕ +2α)+···+ cos(ϕ + nα) = sin (n +1)α 2 cos  ϕ + nα 2  sin α 2 · Gia ’ i. 1) D˘a . t S = cos π 9 + cos 3π 9 + ···+ cos 7π 9 , T = sin π 9 + sin 3π 9 + ···+ sin 7π 9 , z = cos π 9 + i sin π 9 . 1.4. Biˆe ’ udiˆe ˜ nsˆo ´ ph´u . cdu . ´o . ida . ng lu . o . . ng gi´ac 29 Khi d´o S + iT = z + z 3 + z 5 + z 7 = z(1 − z 8 ) 1 −z 2 = z − z 9 1 −z 2 = z +1 1 −z 2 = 1 1 − z = 1  1 −cos π 9  − i sin π 9 =  1 −cos π 9  + i sin π 9  1 −cos π 9  2 + sin 2 π 9 = 1 2 + sin π 9 2  1 −cos π 9  · Do d ´o S = 1 2 · 2) Tu . o . ng tu . . nhu . trong 1) ta k´yhiˆe . u S = cos ϕ + cos(ϕ + α)+···+ cos(ϕ + nα), T = sin ϕ + sin(ϕ + α)+···+ sin(ϕ + nα), z = cos α + i sin α, c = cos ϕ + i sin ϕ. Khi d ´o S + iT = c + cz + ···+ cz n = c(1 − z n+1 ) 1 −z = (cos ϕ + i sin ϕ)[1 −cos(n +1)α − i sin(n +1)α] 1 − cos α −i sin α = (cos ϕ + i sin ϕ)2 sin (n +1)α 2  cos (n +1)α − π 2 + i sin (n +1)α − π 2  2 sin α 2  cos α − π 2 + i sin α −π 2  = sin (n +1)α 2 cos  ϕ + nα 2  sin α 2 + sin (n +1)α 2 sin  ϕ + nα 2 sin α 2 i. T`u . d´o so s´anh phˆa ` n thu . . c v`a phˆa ` na ’ o ta thu du . o . . ckˆe ´ t qua ’ .  B˘a ` ng phu . o . ng ph´ap tu . o . ng tu . . ta c´o thˆe ’ t´ınh c´ac tˆo ’ ng da . ng a 1 sin b 1 + a 2 sin b 2 + ···+ a n sin b n , a 1 cos b 1 + a 2 cos b 2 + ···+ a n cos b n 30 Chu . o . ng 1. Sˆo ´ ph´u . c nˆe ´ u c´ac acgumen b 1 ,b 2 , ,b n lˆa . pnˆen cˆa ´ psˆo ´ cˆo . ng c`on c´ac hˆe . sˆo ´ a 1 ,a 2 , ,a n lˆa . p nˆen cˆa ´ psˆo ´ nhˆan. V´ı d u . 5. T´ınh tˆo ’ ng 1) S n =1+a cos ϕ + a 2 cos 2ϕ + ···+ a n cos nϕ; 2) T n = a sin ϕ + a 2 sin 2ϕ + ···+ a n sin nϕ. Gia ’ i. Ta lˆa . pbiˆe ’ uth´u . c S n + iT n v`a thu du . o . . c Σ=S n + iT n =1+a(cos ϕ + i sin ϕ)+a 2 (cos 2ϕ + i sin 2ϕ)+ + a n (cos nϕ + i sin nϕ). D˘a . t z = cos ϕ + i sin ϕ v`a ´ap du . ng cˆong th´u . c Moivre ta c´o: Σ=1+az + a 2 z 2 + ···+ a n z n = a n+1 z n+1 − 1 az − 1 (nhˆan tu . ’ sˆo ´ v`a mˆa ˜ usˆo ´ v´o . i a z −1) = a n+2 z n − a n+1 z n+1 − a 2 +1 a 2 − a  z + 1 z  +1 (do z + 1 z = 2 cos ϕ) = a n+2 (cos nϕ + i sin nϕ) −a n+1 [cos(n +1)ϕ + i sin(n +1)ϕ] a 2 − 2a cos ϕ +1 + −a cos ϕ + ai sin ϕ +1 a 2 − 2a cos ϕ +1 = a n+2 cos nϕ − a n+1 cos(n +1)ϕ − a cos ϕ +1 a 2 − 2a cos ϕ +1 + + i a n+2 sin nϕ −a n+1 sin(n +1)ϕ + a sin ϕ a 2 − 2a cos ϕ +1 · B˘a ` ng c´ach so s´anh phˆa ` n thu . . c v`a phˆa ` na ’ otathudu . o . . c c´ac kˆe ´ t qua ’ cˆa ` n d u . o . . c t´ınh. V´ı d u . 6. 1) Biˆe ’ udiˆe ˜ n tg5ϕ qua tgϕ. 1.4. Biˆe ’ udiˆe ˜ nsˆo ´ ph´u . cdu . ´o . ida . ng lu . o . . ng gi´ac 31 2) Biˆe ’ udiˆe ˜ n tuyˆe ´ n t´ınh sin 5 ϕ qua c´ac h`am sin cu ’ a g´oc bˆo . icu ’ a ϕ. 3) Biˆe ’ udiˆe ˜ n cos 4 ϕ v`a sin 4 ϕ·cos 3 ϕ qua h`am cosin cu ’ a c´ac g´oc bˆo . i. Gia ’ i. 1) V`ı tg5ϕ = sin 5ϕ cos 5ϕ nˆen ta cˆa ` nbiˆe ’ udiˆe ˜ n sin 5ϕ v`a cos 5ϕ qua sin ϕ v`a cosϕ. Theo cˆong th´u . c Moivre ta c´o cos 5ϕ + i sin 5ϕ = (cos ϕ + i sin ϕ) 5 = sin 5 ϕ +5i cos 4 ϕ sin ϕ − 10 cos 3 ϕ sin 2 ϕ −10i cos 2 ϕ sin 3 ϕ + 5 cos ϕ sin 4 ϕ + i sin 5 ϕ. T´ach phˆa ` n thu . . c v`a phˆa ` na ’ o ta thu d u . o . . cbiˆe ’ uth´u . cd ˆo ´ iv´o . i sin 5ϕ v`a cos 5ϕ v`a t`u . d´o tg5ϕ = 5 cos 4 ϕ sin ϕ −10 cos 2 ϕ sin 3 ϕ + sin 5 ϕ cos 5 ϕ −10 cos 3 ϕ sin 2 ϕ + 5 cos ϕ sin 4 ϕ (chia tu . ’ sˆo ´ v`a mˆa ˜ usˆo ´ cho cos 5 ϕ) = 5tgϕ −10tg 3 ϕ +tg 5 ϕ 1 − 10tg 2 ϕ + 5tg 4 ϕ · 2) D˘a . t z = cos ϕ + i sin ϕ. Khi d´o z −1 = cos ϕ − i sin ϕ v`a theo cˆong th´u . c Moivre: z k = cos kϕ + i sin kϕ, z −k = cos kϕ − i sin kϕ. Do d´o cos ϕ = z + z −1 2 , sin ϕ = z − z −1 2i z k + z −k = 2 cos kϕ, z k −z −k =2i sin kϕ. ´ Ap du . ng c´ac kˆe ´ t qua ’ n`ay ta c´o sin 5 ϕ =  z − z −1 2i  5 = z 5 − 5z 3 +10z − 10z −1 +5z −3 −z −5 32i = (z 5 − z −5 ) −5(z 3 − z −3 )+10(z − z −1 ) 32i = 2i sin 5ϕ −10i sin 3ϕ +20i sin ϕ 32i = sin 5ϕ −5 sin 3ϕ + 10 sin ϕ 16 · 32 Chu . o . ng 1. Sˆo ´ ph´u . c 3) Tu . o . ng tu . . nhu . trong phˆa ` n 2) ho˘a . c gia ’ i theo c´ach sau d ˆay 1 + cos 4 ϕ =  e iϕ + e −iϕ 2  4 = 1 16  e 4iϕ +4e 2iϕ +6+4e −2iϕ + e −4iϕ  = 1 8  e 4ϕi + e −4ϕi 2  + 1 2  e 2ϕi + e −2ϕi 2  + 3 8 = 3 8 + 1 2 cos 2ϕ + 1 8 cos 4ϕ. 2 + sin 4 ϕ cos 3 ϕ =  e ϕi − e −ϕi 2i  4  e ϕi + e −ϕi 2  3 = 1 128  e 2ϕi − e −2ϕi  3  e ϕi −e −ϕi  = 1 128  e 6ϕi − 3e 2ϕi +3e −2ϕi − e −6ϕi  e ϕi −e −ϕi  = 1 128  e 7ϕi −e 5ϕi − 3e 3ϕi +3e ϕi +3e −ϕ i − 3e −3ϕi − e −5ϕi + e −7ϕi  = 3 64 cos ϕ − 3 64 cos 3ϕ − 1 64 cos 5ϕ − 1 64 cos 7ϕ.  V´ı d u . 7. 1) Gia ’ i c´ac phu . o . ng tr`ınh 1 + (x +1) n −(x − 1) n =0 2 + (x + i) n +(x − i) n =0, n>1. 2) Ch´u . ng minh r˘a ` ng mo . i nghiˆe . mcu ’ aphu . o . ng tr`ınh  1+ix 1 −ix  n = 1+ai 1 −ai ,n∈ N,a∈ R d ˆe ` u l`a nghiˆe . m thu . . c kh´ac nhau. Gia ’ i. 1) Gia ’ iphu . o . ng tr`ınh 1 + Chia hai vˆe ´ cu ’ aphu . o . ng tr`ınh cho (x −1) n ta du . o . . c  x +1 x −1  n =1⇒ x +1 x − 1 = n √ 1=cos 2kπ n + i sin 2kπ n = ε k , k =0, 1, ,n− 1. 1.4. Biˆe ’ udiˆe ˜ nsˆo ´ ph´u . cdu . ´o . ida . ng lu . o . . ng gi´ac 33 T`u . d ´o suy r˘a ` ng x +1=ε k (x −1) ⇒ x(ε k − 1)=1+ε k . Khi k =0⇒ ε 0 = 1. Do d´ov´o . i k =0phu . o . ng tr`ınh vˆo nghiˆe . m. V´o . i k = 1,n−1 ta c´o x = ε k +1 ε k − 1 = (ε k + 1)(ε k −1) ε k − 1)(ε k −1) = ε k ε k + ε k − ε k − 1 ε k ε k − ε k − ε k − 1 = −2i sin 2kπ n 2 −2 cos 2kπ n = −i sin 2kπ n 1 −cos 2kπ n = icotg kπ n ,k=1, 2, ,n− 1. 2 + C˜ung nhu . trˆen, t`u . phu . o . ng tr`ınh d˜a cho ta c´o  x + i x − i  n = −1 ⇐⇒ x + i x −i = n √ −1=cos π +2kπ n + i sin π +2kπ n hay l`a x + i x −i = cos (2k +1)π n + i sin (2k +1)π n = cos ψ + i sin ψ,ψ= (2k +1)π n · Ta biˆe ´ ndˆo ’ iphu . o . ng tr`ınh: x + i x −i − 1=cosψ + i sin ψ − 1 ⇔ 2i x −i =2i sin ψ 2 cos ψ 2 −2 sin 2 ψ 2 ⇔ 1 x −i = sin ψ 2  cos ψ 2 − 1 i sin ψ 2  = sin ψ 2  cos ψ 2 + i sin ψ 2  . 34 Chu . o . ng 1. Sˆo ´ ph´u . c T`u . d ´o suy ra x −i = 1 sin ψ 2  cos ψ 2 + i sin ψ 2  = cos ψ 2 − i sin ψ 2 sin ψ 2 = cotg ψ 2 − i. Nhu . vˆa . y x −i = cotg ψ 2 − i ⇒ x = cotg ψ 2 = cotg (2k +1)π 2n ,k= 0,n− 1. 2) Ta x´et vˆe ´ pha ’ icu ’ aphu . o . ng tr`ınh d ˜a cho. Ta c´o    1+ai 1 −ai    =1⇒ 1+ai 1 − ai = cos α + i sin α v`a t `u . d ´o 1+xi 1 − xi = n  1+ai 1 −ai = cos α +2kπ n + i sin α +2kπ n ,k= 0,n− 1. T`u . d ´onˆe ´ ud˘a . t ψ = α +2kπ n th`ı x = cos ψ − 1+i sin ψ i[cos ψ +1+i sin ψ] =tg ψ 2 =tg α +2kπ 2n ,k= 0,n− 1. R˜o r`ang d ´o l`a nh˜u . ng nghiˆe . m thu . . c kh´ac nhau.  V´ı d u . 8. Biˆe ’ udiˆe ˜ n c´ac sˆo ´ ph´u . csaudˆay du . ´o . ida . ng m˜u: 1) z = (− √ 3+i)  cos π 12 −i sin π 12  1 −i · 2) z =  √ 3+i. 1.4. Biˆe ’ udiˆe ˜ nsˆo ´ ph´u . cdu . ´o . ida . ng lu . o . . ng gi´ac 35 Gia ’ i. 1) D˘a . t z 1 = − √ 3+i, z 2 = cos π 12 − i sin π 12 , z 3 =1− i v`a biˆe ’ udiˆe ˜ n c´ac sˆo ´ ph´u . cd´odu . ´o . ida . ng m˜u. Ta c´o z 1 =2e 5π 6 i ; z 2 = cos π 12 − i sin π 12 = cos  − π 12  + i sin  − π 12  = e − π 12 i ; z 3 = √ 2e − π 4 i . T`u . d´othudu . o . . c z = 2e 5π 6 i · e − π 12 i √ 2e − π 4 i = √ 2e iπ . 2) Tru . ´o . chˆe ´ tbiˆe ’ udiˆe ˜ nsˆo ´ ph´u . c z 1 = √ 3+i du . ´o . ida . ng m˜u. Ta c´o |z 1 | =2; ϕ = arg( √ 3+i)= π 6 , do d ´o √ 3+i =2e π 6 i .T`u . d ´othudu . o . . c w k = 4  √ 3+i = 4 √ 2e i ( π 6 +2kπ) 4 = 4 √ 2e i (12k+1)π 24 ,k= 0, 3.  V´ı d u . 9. T´ınh c´ac gi´a tri . 1) c˘an bˆa . c3: w = 3 √ −2+2i 2) c˘an bˆa . c4: w = 4 √ −4 3) c˘an bˆa . c5: w = 5  √ 3 −i 8+8i . Gia ’ i. Phu . o . ng ph´ap tˆo ´ t nhˆa ´ td ˆe ’ t´ınh gi´a tri . c´ac c˘an th´u . cl`abiˆe ’ u diˆe ˜ nsˆo ´ ph´u . cdu . ´o . idˆa ´ u c˘an du . ´o . ida . ng lu . o . . ng gi´ac (ho˘a . cda . ng m˜u) rˆo ` i ´ap du . ng c´ac cˆong th´u . ctu . o . ng ´u . ng. 1) Biˆe ’ udiˆe ˜ n z = −2+2i du . ´o . ida . ng lu . o . . ng gi´ac. Ta c´o r = |z| = √ 8=2 √ 2; ϕ = arg(−2+2i)= 3π 4 · 36 Chu . o . ng 1. Sˆo ´ ph´u . c Do d´o w k = 3  √ 8  cos 3π 4 +2kπ 3 + i sin 3π 4 +2kπ 3  ,k= 0, 2. T`u . d´o w 0 = √ 2  cos π 4 + i sin π 4  =1+i, w 1 = √ 2  cos 11π 12 + i sin 11π 12  , w 2 = √ 2  cos 19π 12 + i sin 19π 12  . 2) Ta c´o −4 = 4[cos π + i sin π] v`a do d ´o w k = 4 √ 4  cos π +2kπ 4 + i sin π +2kπ 4  ,k= 0, 3. T`u . d ´o w 0 = √ 2  cos π 4 + i sin π 4  =1+i, w 1 = √ 2  cos 3π 4 + i sin 3π 4  = −1+i, w 2 = √ 2  cos 5π 4 + i sin 5π 4  = −1 − i, w 3 = √ 2  cos 7π 4 + i sin 7π 4  =1−i. 3) D ˘a . t z = √ 3 −i 8+8i · Khi d ´o |z| = √ 3+1 √ 64 + 64 = 1 4 √ 2 . Ta t´ınh argz.Tac´o argz = arg( √ 3 −i) − arg(8 + 8i)=− π 6 − π 4 = − 5π 12 · [...]... θ (DS 2 cos cos + i sin v´.i 0 θ < π; o 2 2 2 3π − θ 3π − θ θ cos + i sin v´.i π θ < 2 ) o 2 cos 2 2 2 8) − sin α + i (1 + cos α) π+θ π+θ θ cos + i sin v´.i 0 θ < π; o (DS 2 cos 2 2 2 3π + θ 3π + θ θ cos + i sin v´.i π θ < 2 ) o 2 cos 2 2 2 3 T´ ınh: √ π π 10 0 1 3 1) cos − i sin ) (DS − − i 6 6 2 2 12 4 2) √ (DS 21 2 ) 3+i √ ( 3 + i)6 3) (DS −3, 2) ( 1 + i)8 − (1 + i)4 √ √ (−i − 3 )15 (−i + 3 )15 4)... ıch a 1) z 3 − 6z 2 + 11 z − 6 2) 6z 4 − 11 z 3 − z 2 − 4 (DS (z − 1) (z − 2) (z − 3)) √ 1 i 3 z− 2 2 2 3) 3z 4 − 23 z 2 − 36 z + i√ ) (DS 3(z − 3)(z + 3) z − i √ 3 3 n (DS (z − ε0 )(z − 1) · · · (z − εn 1 ), 4) z − 1 2kπ 2kπ εk = cos + i sin , k = 0, n − 1) n n (DS (z − 1 − i)(z − 1 + i)(z + 1 − i)(z + 1 + i)) 5) z 4 + 4 2 (DS 6(z − 2) z + 3 √ 1+ i 3 z− 2 6) z 4 + 16 √ √ √ √ (DS (z − 2 (1 + i))(z − 2 (1 −... 4) 2 2 6 6 √ 5π 5π − 3 1 + i (DS cos + i sin ) 5) 2 2 6 6 √ 5π 5π 3 1 (DS cos + i sin ) 6) − i 2 2 3 3 √ 4π 4π 3 1 (DS cos + i sin ) 7) − − i 2 2 3 3 √ √ 23 π 23 π + i sin ) (DS 2 2 + 3 cos 8) 2 + 3 − i 12 12 √ √ 19 π 19 π + i sin ) (DS 2 2 − 3 cos 9) 2 − 3 − i 12 12 ’ ˜ a o u ´ o a 2 Biˆu diˆn c´c sˆ ph´.c sau dˆy du.´.i dang lu.o.ng gi´c e e a 1) − cos ϕ + i sin ϕ (DS cos(π − ϕ) + i sin(π − ϕ)) π π 2) ... (DS −64i) (1 − i )20 (1 + i )20 5) (1 + i )10 0 (1 − i)96 + (1 + i)96 (DS 2) (1 + icotgϕ)5 (DS cos(π − 10 ϕ) + i sin(π − 10 ϕ)) 1 − icotgϕ)5 √ (1 − i 3)(cos ϕ + i sin ϕ) 7) 2 (1 − i)(cos ϕ − i sin ϕ) √ π π 2 cos 6ϕ − + i sin 6ϕ − ) (DS 2 12 12 6) ’ ˜ o u 1. 4 Biˆu diˆn sˆ ph´.c du.´.i dang lu.o.ng gi´c e e ´ o a √ (1 + i 3)3n 8) (1 + i)4n (DS 2) 1 1 ` a 4 Ch´.ng minh r˘ng z + = 2 cos ϕ ⇒ z n + n = 2 cos nϕ... + 2 (1 + i))(z + 2 (1 − i))) 7) z 4 + 8z 3 + 8z − 1 √ √ 17 )(z + 4 + 17 )) √ √ 1 i 7 1+ i 7 3 8) z + z + 2 z− ) (DS (z + 1) z − 2 2 ´ ´ u u a e o o a a 7 Phˆn t´ c´c da th´.c trˆn tru.`.ng sˆ thu.c th`nh c´c da th´.c bˆt a ıch a `.ng d´ ’ o kha quy trˆn c`ng tru o e u (DS (x + 1) (x2 − x + 2) ) 1) x3 + x + 2 √ √ (DS (x2 − 2x 2 + 4)(x2 + 2 2x + 4)) 2) x4 + 16 √ √ 3) x4 + 8x3 + 8x − 1 (DS (x2 + 1) (x + 4 − 17 )(x... Nˆu n > 1 th` εn = 1 v` do d´ e ı S= 1 − εn = 0 1 ε 1 − εn · 1 ε ´ Chu.o.ng 1 Sˆ ph´.c o u 38 ’ ’ ` ınh a 2) Ta k´ hiˆu tˆng cˆn t´ l` S Ta x´t biˆu th´.c y e o a e e u (1 − ε)S = S − εS = 1 + 2 + 3 2 + · · · + nεn 1 − ε − 2 2 − · · · − (n − 1) εn 1 − nεn = 1 + ε + 2 + · · · + εn 1 −nεn = −n 0(ε =1) v` εn = 1 ı a Nhu vˆy (1 − ε)S = −n → S = −n 1 ε ´ nˆu ε = 1 e ´ Nˆu ε = 1 th` e ı S = 1 + 2 + ···... 2 .1 Da th´.c u 47 - ’ ’ o a e Dinh l´ 2 .1. 2 Gia su da th´.c Q(x) c´ c´c nghiˆm thu.c b1, b2 , , bm y u i bˆi tu.o.ng u.ng 1, 2, , βm v` c´c c˘p nghiˆm ph´.c liˆn ho.p a1 ´ v´ o o a a a e u e v` a1 , a2 v` a2 , , an v` an v´.i bˆi tu.o.ng u.ng 1 , 2 , , λn Khi d´ a a a o o o ´ Q(x) = (x − b1 ) 1 (x − b2 ) 2 · · · (x − bm )βm (x2 + p1 x + q1) 1 × × (x2 + p2 x + q2) 2 · · · (x2... εn 1 , 2 = εn 2 , 2 .1 Da th´.c u 53 M˘t kh´c a a (x − εk )(x − εk ) = x2 − (εk + εk )x + εk εk = x2 − x · 2 cos 2kπ + 1 n Do d´ o  n 1  2 2kπ  (x − 1) ´ ´ x2 − x · 2 cos + 1 nˆu n l` sˆ le, e a o ’    n k =1  n 2 2 xn − 1 = 2kπ (x − 1) (x + 1) +1 x2 − x · 2 cos   n  k =1    ´ ´ a nˆu n l` sˆ ch˘ n e a o ˜ ` ˆ BAI TAP ` ´ ’ a e u a o 1 Ch´.ng minh r˘ng sˆ z0 = 1 + i l` nghiˆm cua da... k = β0 (1 + εk + εk + · · · + εk ) 1 2 n 1 εk = m cos 2mπ 2mπ + i sin n n (n 1) k k = β0 1 + εk + ε2k + · · · + 1 1 1 k = cos 2 2 + i sin n n mk ’ ´ ´ ´ ´ u a a a o a a o a e 1 Biˆu th´.c trong dˆu ngo˘ c vuˆng l` cˆp sˆ nhˆn Nˆu εk = 1, t´.c l` e u ´ k khˆng chia hˆt cho n th` o e ı k S = β0 1 − εnk 1 k 1 1 = β0 = 0 (v` εn = 1) ı 1 k 1 − 1 1 − εk 1 ’ ˜ o u 1. 4 Biˆu diˆn sˆ ph´.c du.´.i dang lu.o.ng... 4k)π , k = 0, n − 1) (DS x = −cicotg 4n 4) (x + ci)n − (cos α + i sin α)(x − ci)n = 0, α = 2kπ α + 2kπ (DS x = −cicotg , k = 0, n − 1) 2n 15 T´ ınh Dn (x) = 1 1 + cos x + cos 2x + · · · + cos nx 2 2 2n + 1 1 sin 2 x (DS Dn (x) = ) 2 2 sin x 2 ’u diˆn cos 5x v` sin 5x qua cos x v` sin x ˜ 16 1) Biˆ e e a a 2 2 2) T´nh cos ı v` sin a 5 5 (DS 1) cos 5x = cos5 x − 10 cos3 x sin2 x + 5 cos x sin4 . d´o w k = 3  √ 8  cos 3π 4 +2kπ 3 + i sin 3π 4 +2kπ 3  ,k= 0, 2. T`u . d´o w 0 = √ 2  cos π 4 + i sin π 4  =1+ i, w 1 = √ 2  cos 11 π 12 + i sin 11 π 12  , w 2 = √ 2  cos 19 π 12 + i sin 19 π 12  . 2) Ta c´o −4 = 4[cos. sin π 6  10 0 (DS. − 1 2 − i √ 3 2 ) 2)  4 √ 3+i  12 (DS. 2 12 ) 3) ( √ 3+i) 6 ( 1+ i) 8 − (1 + i) 4 (DS. −3, 2) 4) (−i − √ 3) 15 (1 −i) 20 + (−i + √ 3) 15 (1 + i) 20 (DS. −64i) 5) (1 + i) 10 0 (1 −i) 96 + (1+ i) 96 (DS e −4ϕi 2  + 1 2  e 2 i + e 2 i 2  + 3 8 = 3 8 + 1 2 cos 2 + 1 8 cos 4ϕ. 2 + sin 4 ϕ cos 3 ϕ =  e ϕi − e −ϕi 2i  4  e ϕi + e −ϕi 2  3 = 1 128  e 2 i − e 2 i  3  e ϕi −e −ϕi  = 1 128  e 6ϕi −