Universitas Jember Jurusan Matematika - FMIPA
MAM 1516 Analisis Kompleks
Deadline: Wednesday, 29–10–2014; 23:55 Tugas 2 – Template Jawaban
Nama Kelompok: Group J Nama Anggota:
1. Darul Afandi (111810101041)
2. Wahyu Nikmatus Sholihah (121810101010) 3. Irawati NIM (121810101021)
4. Kiki Kurdianto (121810101041)
5. Reyka Bella Desvandai (121810101080)
1 Nama Anggota 1:Darul Afandi (111810101041) Jawaban soal No 40.
———————————————————————————————————————- Soal no.40:
Hitunglah Hitung H (5z4 − z3 + 2)dz disekeliling (a) Lingkaran |z | = 1, (b) bujur sangkar dengan titik-titik sudut (0,1),(1,0),(1,1) dan (1,0), (c) kurva yang dibatasi parabola y = x2 dari (0,0) ke (1,1) dan y = x2 dari (1,1) ke (0,0)
Solusi:
a. Hitung H (5z4− z3+ 2)dz disekeliling Lingkaran |z | = 1
Penyelesaiaan:
|z| = 1 px2+ y2 = 1
x2+ y2 = 1 → r = 1
x = r cos θ = cos θ → dx = − sin θ dθ y = r sin θ = sin θ → dy = cos θ dθ
0 ≤ θ ≤ 2π
z4 = re4πθ
z4 = 1(cos 4θ + i sin4θ) z3 = (cos 3θ + i sin3θ)
= I
c
(5z4− z3+ 2)dz
= Z 2pi
0
5(cos 4θ + isin4θ) − (cos 3θ + i sin3θ) + 2(dx + idy)
= Z 2pi
0
(5 cos 4θ + 5i sin 4θ − cos 3θ − i sin 3θ + 2)(− sin θ dθ + i cos θ dθ)
= Z 2pi
0
−5 cos 4θ sin θ + 5i cos 4θ cos θ − 5i sin 4θ sin θ − 5 sin 4θ cos θ + cos 3θ sin θ
− i cos 3θ cos θ + i sin 3θ sin θ + sin 3θ cos θ − 2 sin θ + 2i cos θ dθ Z
−5
2(sin 5θ − sin 3θ) −5
2(sin 5θ + sin 3θ) + 1
2(sin 4θ − sin 2θ) + 1
2(sin 4θ + sin 2θ) + 2 cos θ dθ + i
Z 5
2(cos 5θ + cos 3θ) − 5
2(cos 5θ − cos 3θ) − 1
2(cos 4θ + cos 2θ)
− 1
2(cos 4θ − cos 2θ) + 2 sin θ dθ
= Z
−5
2sin 5θ +5
2sin 3θ − 5
2sin 5θ − 5
2sin 3θ +1
2sin 4θ −1
2sin 2θ +1
2sin 4θ +1 2sin 2θ + 2 cos θ dθ + i
Z 5
2cos 5θ + 5
2cos 3θ − 5
2cos 5θ + 5
2cos 3θ − 1
2cos 4θ −1 2cos 2θ
− 1
2cos 4θ − 1
2cos 2θ + 2 sin θ dθ
= Z
−5 sin 5θ + sin 4θ + 2 cos θ dθ + i Z
5 cos 3θ − 4 cos 4θ − cos 2θ + 2 sin θ dθ
= [5.1
5cos 5θ −1
4cos 4θ + 2 sin θ + i(5.1
3sin 3θ − 1
4sin 4θ − 1
2sin 2θ − 2 cos θ)]2θ0
= [cos 5θ − 1
4cos 4θ + 2 sin θ + i(5
3sin 3θ − 1
4sin 2θ − 2 cos θ)]2π0
= [cos 10π − 1
4cos 8π + 2 sin 2π + i(5
3sin 6π −1
4sin 8π − 1
2sin 4π − 2 cos 2π)]
− [cos 0 − 1
4cos 0 + 2 sin 0 + i(5
3sin 0 − 1
4sin 0 − 1
2sin 0 − 2 cos 0)]
= [1 − 1
4 + 0 + i(0 − 0 − 0 − 2)] − [1 −1
4 + 0 + i(0 − 0 − 0 − 2)]
= 3
4− 2i) − (3 4 − 2i)
= 3
4− 2i − 3 4+ 2i
b. Bujursangkar dengan titik-titik sudut (0, 0), (1, 0), (1, 1), dan (0, 1) C1=Pada titik (0, 0) ke (1, 0) memiliki persamaan y = 0, dy = 0
Z 1,0 0,0
(5z4− z3+ 2)dz = Z 1
0
5((x + iy)4− (x + iy)3+ 2)d(x + iy)
= Z 1
0
5((x + iy)4− (x + iy)3+ 2)d(x + iy)
= Z 1
0
(5(x + 0)4− (x + 0)3 + 2)dx
= Z 1
0
5x4 − x3 + 2dx
= [x5− (1
4)4+ 2x]10
= [1 − (1 4) + 2]
= 11 4
C2=Pada titik (1, 0) ke (1, 1) memiliki persamaan x = 1, dx = 0
Z 1,1 1,0
(5z4− z3+ 2)dz = Z 1
0
5((x + iy)4− (x + iy)3+ 2)d(x + iy)
= Z 1
0
5((x + iy)4− (x + iy)3+ 2)d(x + iy)
= Z 1
0
(5(1 + iy)4− (1 + iy)3+ 2)idy
= i Z 1
0
5(1 + 4iy − 6y2− 4iy3+ y4)
− (1 + 3iy − 3y2− iy3) + 2dy
= i Z 1
0
(5 + 20iy − 30y2− 20iy3+ 5y4)
− 1 − 3iy + 3y2+ iy3+ 2dy
= i Z 1
0
(6 + 17iy − 27y2− 19iy3+ 5y4)dy
= i[6y + 17
2 iy2− 9y3−19
4 iy4+ y5]10
= i[(6 + 17
2 i − 9 −19
4 i + 1) − 0]
= i[−2 + 15 4 ]
= −15 4 − 2i
C3=Pada titik (1, 1) ke (1, 0) memiliki persamaan x = 1 dx = 0
Z 1,0 1,1
(5z4− z3+ 2)dz = Z 0
1
5((x + iy)4− (x + iy)3+ 2)d(x + iy)
= Z 0
1
5((x + iy)4− (x + iy)3+ 2)d(x + iy)
= Z 0
1
(5(1 + iy)4− (1 + iy)3+ 2)idy
= i Z 0
1
5(1 + 4iy − 6y2− 4iy3+ y4)
− (1 + 3iy − 3y2− iy3) + 2dy
= i Z 0
1
(5 + 20iy − 30y2− 20iy3+ 5y4)
− 1 − 3iy + 3y2+ iy3+ 2dy
= i Z 0
1
(6 + 17iy − 27y2− 19iy3+ 5y4)dy
= i[6y + 17
2 iy2− 9y3−19
4 iy4+ y5]01
= i[0 − (6 + 17
2 i − 9 −19 4 i + 1)]
= i[2 − 15 4 ]
= 2i + 15 4
C4=Pada titik (1, 0) ke (0, 0) memiliki persamaan y = 0, dy = 0 Z 0,0
1,0
(5z4− z3+ 2)dz
= Z 0
1
(5(x + 0)4− (x + 0)3 + 2)dx
= Z 0
1
5x4 − x3 + 2dx
= [x5− (1
4)4+ 2x]01
= [0 − (1 − (1
4) + 2)]
= −11 4 Penyelesaiannya yang diinginkan adalah :
C1 + C2 + C3 + C4 = (11
4 ) + (−15
4 − 2i) + (2i + 15
4 ) + (−11 4 )
= 0
c. Kurva Pada Parabola y = x2 dari(0, 0) ke (1, 1) dan y2 = x dari (1, 1) ke (0, 0)
1). Lintasan C1, y = x2 → dy = 2x dx I
c
(5z4− z3+ 2) dz
= I
c
[5(x + iy)4− (x + iy)3+ 2] (dx + idy)
= I
c
5(x4+ y4− 6x2y2+ 4ix3y − 4ixy3) − (x3− 3xy2+ 3ix2y − iy3) + 2(dx + i2x dx)
= I
c
5x4+ 5y4− 30x2y2+ 20ix3y − 20ixy3− x3+ 3xy2− 3ix2y + iy3+ 2 (dx + i dy)
= I
c
5x4+ 5(x2)4− 30x2x4+ 20i x3x2− 20ix − x6− x3+ 3xx4− 3ix2x2+ ix6 + 2 (dx + i2x dx)
= I
c
5x4+ 5x8− 30x6+ 20ix5+ 20ix7− x3+ 3x5− 3ix4+ ix6+ 2 (dx + i2x dx)
= Z
5x4+ 5x8− 30x6− x3+ 3x5+ 2 − 40x6+ 40x8+ 6x5− 2x7 dx + i Z
10x5+ 10x9
− 60x7+ 20x5− 20x7− 2x4+ 6x6 − 3x4+ x6+ 4x dx
= Z
45x8− 2x7− 70x6+ 9x5+ 5x4− x3+ 2 dx + i
Z
(10x9− 80x7+ 7x6+ 30x5− 5x4+ 4x) dx
= [45
9 x9− 2
8x8− 70
7 x7+ 9
6x6+ 5
5x5 −1
4x4+ 2 dx + i(10
10x10 −80
8 x8+ 7
7x7 +30
6 x6+ 5
5x5+ 4 2x2)]10
= [5x9 −1
4x8− 10x7+ 3
2x6 + x5− 1
4x4+ 2x + i(x10 − 10x8+ x7 + 5x6− x5+ 2x2)]10
= [5 − 1
4 − 10 + 2
3+ 1 − 1
4 + 2 + i(1 − 10 + 1 + 5 − 1 + 2)] − 0[20 − 1 − 40 + 6 + 4 − 1 + 8 4
+ i(−2)] = −4 4 − 2i
= −1 − 2i
2). Lintasan C2, x = y2 → dx= 2y dy Z 0
1
(5x4+ 5y4− 30x2y2+ 20ix3y − 20ixy3− x3+ 3xy2− 3ix2y + iy3+ 2) (dx + idy)
= Z 0
1
(5y8+ 5y4− 30y8y2 + 20iy6y − 20iy2y3− y6 + 3y2y2 − 3iy4y + iy3+ 2) (2y dy + i dy)
= Z 0
1
(5y8+ 5y4− 30y6+ 20iy7− 20iy5 − y6+ 3y4− 3iy5+ iy3+ 2) (2y dy + i dy)
= Z 0
1
(5y8+ 20iy7− 31y6− 23iy5+ 8y4 + iy3+ 2) (2y dy + i dy)
= Z 0
1
10y9− 20y7− 62y7+ 23y5+ 16y5− y3+ 4y dy + i Z
(5y8+ 40y8− 31y6
− 46y6+ 8y4+ 2y4+ 2 dy
= Z 0
1
10y9− 82y7+ 39y5− y3+ 4y dy + i Z
(5y8+ 40y8− 31y6− 46y6+ 8y4+ 2y4+ 2 dy
= [10
10y10 − 82
8 y8+ 39 6 y6− 1
4y4+ 2y2+ i(45
9 y9 −77
7 y7+10
5 y5+ 2y)]01
= [y10 − 41
4 y8+ 13 2 y6− 1
4y4+ 2y2 + i(5y − 11y7+ 2y5+ 2y)]01
= (1 − 41 4 + 13
2 − 1
4 + 2 + i(5 − 11 + 2 + 2))
= −(4 − 41 + 26 − 1 + 8 + i
4 (−2)) = (−4
4 − 2i) = 1 + 2i
∴ Jadi, C1+ C2
= (−1 − 2i) + (1 + 2i)
= −1 − 2i + 1 + 21
= 0
2 Nama Anggota 2: Wahyu Nikmatus sholihah (121810101010) Jawaban soal no.39
———————————————————————————————————————- Soal No.39
Hitunglah H
C
z2dz disekeliling lingkaran (a) |z| = 1 dan (b) |z − 1| = 1
Jawab (a)
|z| = 1 px2+ y2 = 1 x2+ y2 = 1
x = rcosθ dx = −sinθdθ 0 ≤ θ ≤ 2π
y = rsinθ dy = cosθdθ
H
C
z2dz
=H
C
(x − iy)2(dx + idy)
=H
C
(x2− y2− 2ixy)(dx + idy)
=R2π
0 (cos2θ − sin2θ − 2i cos θ sin θ)(− sin θdθ + i cos θdθ)
=R2π
0 (cos2θ − sin2θ − i sin 2θ)(− sin θdθ + i cos θdθ)
=R2π
0 (− cos2θ sin θ + sin3θ + i sin 2θ sin θ + i sin3θ − i sin2θ cos θ + i sin 2θ cos θ)dθ
=R2π
0 (sin3θ − cos2θ sin θ + sin 2θ cos θ)dθ + iR2π
0 (cos3θ − sin2θ cos θ + sin 2θ sin θ)dθ
= ((− cos θ + 13cos3θ) + (13cos3θ) + (−13cos 3θ − cosθ))|2π0 + i((sin θ − 13 sin3θ) − (13sin3θ) + (23sin 3θ))|2π0
= ([(−1 + 13) + 13 + (−13 − 1)] − [(−1 + 13) + 13 + (−13 − 1)]) − i(0)
= ([−2 + 13] − [−2 + 13]) − 0
= 0 − 0
= 0
Hasil yang didapat adalah 0 (b)
|z − 1| = 1 p(x − 1)2 + y2 = 1 (x − 1)2 + y2 = 1
(x − 1) = rcosθ y = rsinθ
x = rcosθ − 1 dx = −sinθdθ 0 ≤ θ ≤ 2π
y = rsinθ dy = cosθdθ
H
C
z2dz
=H
C
(x − iy)2(dx + idy)
=H
C
(x2− y2− 2ixy)(dx + idy)
=R
C
x2dx −R
C
y2dx − 2iR
C
xydx + iR
C
x2dy − iR
C
y2dy + 2R
C
xydy
=R
C
x2dx −R
C
y2dx + 2R
C
xydy + i(R
C
x2dy −R
C
y2dy − 2R
C
xydx) R
C
x2dx −R
C
y2dx + 2R
C
xydy ... persamaan (1) i(R
C
x2dy −R
C
y2dy − 2R
C
xydx) ... persamaan (2)
Persamaan (1) R
C
x2dx −R
C
y2dx + 2R
C
xydy
=R2π
0 (cos θ + 1)2(− sin θdθ) −R2π
0 sin2θ(− sin θdθ) + 2R2π
0 (cos θ + 1)(sin θ) cos θdθ
=R2π
0 (cos2θ + 2 cos θ + 1)(− sin θdθ) +R2π
0 sin3θdθ + 2R2π
0 (cos θ + 1)(sin θ) cos θdθ
= −R2π
0 (cos2θ sin θ + 2 cos θ sin θ + sin θ)dθ+R2π
0 sin3θdθ+2R2π
0 (cos2θ sin θ + cos θ sin θ)dθ
=R2π
0 (cos2θ sin θ)dθ −R2π
0 sin θdθ +R2π
0 sin3θdθ
= (13cos3θ)|2π0 − (12sin2θ)|2π0 + (− cos θ + 13cos3θ)|2π0
= 0 − 0 + 0
= 0
Persamaan (2) i(R
C
x2dy −R
C
y2dy − 2R
C
xydx)
= i[R2π
0 (cos θ + 1)2(cos θdθ) −R2π
0 sin2(cos θdθ) − 2R2π
0 (cos θ + 1)(sin θ)(− sin θdθ)]
= i[R2π
0 (cos3θ + 2 cos2θ + cos θ)dθ −R2π
0 sin2cos θdθ + 2R2π
0 (sin2θ cos θ + sin2θ)dθ]
= i[R2π
0 (cos3θ + 2 cos2θ + cos θ)dθ +R2π
0 sin2cos θdθ + 2R2π
0 (sin2θ)dθ]
= i[((sin θ + 13sin3θ) + 2(θ2 +12sin θ) + (sin θ))|2π0 + (13sin3θ)2π0 + (θ2 − 12sin θ)2π0 ]
= i[(0 + 2((π + 0) − 0) + 0 + 0 + 2((π − 0) − 0)]
= i[2π + 2π] = 4iπ
Persamaan (1) + Persamaan (2)
= 0 + 4iπ
= 4iπ
Hasil yang didapat adalah 4iπ
3 Nama Anggota 3:Irawati (121810101010) Jawaban soal no 42 dan 46.
———————————————————————————————————————
Soal No.42 Hitunglah R
C
z2dz + z2dz sepanjang kurva yang didefinisikan oleh z2+ 2z z + z2 = (2 − 2i)z + (2 + 2i)z dari titik z=1 ke z=2+2i
Solusi R
C
z2dz + z2dz didefinisikan oleh :
z2+ 2zz + z2 = (2 − 2i)z + (2 + 2i)z dari titik z = 1 ke z = 2 + 2i
Z
C
(x − iy)2d(x + iy) + (x + iy)2d(x − iy)
= Z
C
(x2− 2ixy − y2)(dx + idy) + (x2+ 2ixy − y2)(dx − idy)
= Z
C
(x2− 2ixy − y2)dx + i(x2− 2ixy − y2)dy + (x2+ 2ixy − y2)dx + i(x2+ 2ixy − y2)dy
= Z
C
x2− 2ixy − y2+ x2+ 2ixy − y2)dx + i(x2− 2ixy − y2− x2− 2ixy + y2)dy
= Z
C
(2x2− 2y2)dx + i(−4ixy)dy
= Z
C
(2x2− 2y2)dx + (4xy)dy...P ers(1)
Misal z = x + iy dan z = x − iy, maka:
z2+ 2zz + z2 = (2 − 2i)z + (2 + 2i)z
x2+ 2ixy − y2+ 2(x + iy)(x − 2y) + x2− 2ixy − y2 = (2 − 2i)(x + iy) + (2 + 2i)(x − iy) 2x2− 2y2+ 2x2+ 2y2 = 2x + 2iy − 2ix + 2y + 2x − 2iy + 2ix + 2y
4x2 = 4x + 4y x2 = x + y
y = x2− x...pers(2) dy = (2x − 1)dx...pers(3) Persamaan (2) dan (3) disubtitusikan ke persamaan (1):
R
C
(2x2− 2(x2 − x))dx + 4x(x2− x)(2x − 1)dx
=R
C
2x − 2(x4− 2x3+ x2)dx + 4x(2x3− 3x2+ x)dy
=R
C
(2x2 − 2x2+ 4x3− 2x2)dx + (8x4− 12x3+ 4x2)dx
=R2
0 (6x4− 8x3+ 4x2)dx
= 65x5− 2x4+43x3]21
= 65(31) − 2(15) +43(7)
= 24815
Jadi R
C
z2dz + z2dz sepanjang kurva yang didef inisikan oleh z2+ 2zz + z2 = (2 − 2i)z + (2 + 2i)z dari titik z = 1 ke z = 2 + 2i adalah 24815
Soal No.46
Hitunglah H (5x + 6y − 3)dx(3x − 4y + 2)dy di sekeliling suatu segitiga di bidang xy dengan titik sudut (0,0), (4,0) dan (4,3)
Solusi
H (5x + 6y − 3)dx(3x − 4y + 2)dy menggunakan teorema Green :
pada titik(0, 0)(4, 3) y − y1
y2 − y1 = x − x1 x2 − x1 y − 0
3 − 0 = x − 0 4 − 0 (y − 0)4 = (x − 0)3
4y = 3x y = 3
4x
I
P dx + Qdy = Z Z
R
(dQ dx − dP
dy)dxdy
= Z Z
R
( d
dx(3x − 4y + 2) − d
dy(5x + 6y − 3))dxdy
= Z Z
R
(3 − 6)dxdy
= Z Z
R
(−3y)dydx
= Z 4
0
Z 4y3
0
(−3y)dydx
= Z 4
0
(−3y)]
3 4
0ydx
= Z 4
0
(−3(3
4y) − 0)dx
= Z 4
0
(−9 4x)dx
= −9 8x2]40
= −9
8.42 − 0
= −9 8.16
= −18
Jadi H (5x + 6y − 3)dx(3x − 4y + 2)dy di sekeliling suatu segitiga di bidang xy dengan titik sudut (0, 0) , (4, 0) dan(4, 3) adalah − 18
4 Nama Anggota 4:Kiki Kurdianto (121810101041) Jawaban soal no 50 dan 72
———————————————————————————————————————
Soal No 50:
Periksa teorema green di bidang untuk H
C
x2ydx + y3− xy2dy dimana C batas daerah yang dikelilingi suatu lingkaran x2+ y2 = 4, x2+ y2 = 16.
Solusi:
Perhitungan dengan Teorema Green Misal
P = x2y maka ∂P/∂y = x2
Q = y3− xy2 maka ∂Q/∂x = −y2
I
C
x2ydx + y3− xy2dy = Z Z
∂Q
∂x − ∂P
∂xdA
= Z 2π
0
Z 2 4
(−y2− x2)rdrdΘ
= Z 2π
0
Z 2 4
(−r2sin2Θ − r2cos2Θ)rdrdΘ
= Z 2π
0
((−r4/4) sin2Θ − (r4/4) cos2Θ)]24dΘ
= Z 2π
0
(−4 sin2Θ − 4 cos2Θ − (−64 sin2Θ − 64 cos2Θ))dΘ
= Z 2π
0
(60 sin2Θ + 60 cos2Θ)dΘ
= 60 Z 2π
0
(sin2Θ + cos2Θ)dΘ
= 60 Z 2π
0
(1)dΘ
= 60.Θ]2π0
= 60.2π − (−60).0
= 120π
Perhitungan dengan menggunakan Integral Garis Untuk x2+ y2 = 16
x = 4cost → maka∂x = −4 sin tdt y = 4sint → maka∂y = 4 cos tdt
dengan 0 < t < 2π Maka,
Z 2π 0
((4 cos t)2(4 sin t))d(4 cos t) + ((4 sin t)3 − (4 cos t)(4 sin t)2)d(4 sin t)dx
= Z 2π
0
((16 cos2t.4 sin t)(−4 sin t))d(t) + ((64 sin3t − 64 sin2t cos t)(4 cos t)2)d(t)dx
= Z 2π
0
((−256 cos2t sin2t + 256 cos t sin3t − 256 cos2t sin2t))d(t)
= Z 2π
0
((−512 cos2t sin2t + 256 cos t sin3t))d(t)
= Z 2π
0
(−512 cos2t sin2t)d(t) + Z 2π
0
(256 cos t sin3t)d(t)
Z 2π 0
(−512 cos2t sin2t)d(t) = −512 Z 2π
0
(cos2t sin2t)d(t)
= −512 Z 2π
0
(1 + cos 2t 2
1 − cos 2t 2 )d(t)
= −512 4
Z 2π 0
((1 + cos 2t)(1 − cos 2t))d(t)
= −128 Z 2π
0
((1 − cos 2t + cos 2t − cos22t))d(t)
= −128 Z 2π
0
(1 − cos22t)d(t)
= −128 Z 2π
0
(1 − cos22t)d(t)
= −128 Z 2π
0
(1 −1
2(1 + cos 4t))d(t)
= −128 Z 2π
0
(1 −1
2− cos 4t 2 )d(t)
= −128 Z 2π
0
(1
2 −cos 4t 2 )d(t)
= −128[
Z 2π 0
(1
2)dt − 1 2
Z 2π 0
(cos 4t)dt]
= −128([1
2t]2π0 − 1 2
Z 2π 0
(cos w)dw 4 )
= −128([1
2t]2π0 − [1
8sin 4t]2π0 )
= −128(π − (1
8sin 8π − 1
8sin 0))
= −128π + 16sin8π
= −128π
Z 2π 0
(256 cos t sin3t)d(t) = 256 Z 2π
0
(cos t sin3t)d(t)
= 256 Z 2π
0
((1 − cos2t)(cos t)(sin t))d(t)
= 256 Z 2π
0
(cos t − cos3t)(sin t))d(t)
= 256[
Z 2π 0
(cos t sin t)dt − Z 2π
0
(cos3t sin t)dt]
= 256[
Z 2π 0
(u sin t) du
− sin t− Z 2π
0
(u3sin t) du
− sin t]
= 256([−1
2u2+1 4u4]2π0 ]
= 256([−1
2cos2t + 1
4cos4t]2π0 ]
= 256[(−1 2 +1
4) − (−1 2 +1
4)]
= 0 Sehingga,R2π
0 ((4 cos t)2(4 sin t))d(4 cos t) + ((4 sin t)3− (4 cos t)(4 sin t)2)d(4 sin t)dx = −128π Untuk x2+ y2 = 4
x = 2 cos t → maka∂x = −2 sin tdt y = 2 sin t → maka∂y = 2 cos tdt dengan −2π < t < 0
Maka,
Z 0
−2π
((2 cos t)2(2 sin t))d(2 cos t) + ((2 sin t)3− (2 cos t)(2 sin t)2)d(2 sin t)dx
= Z 0
−2π
((4 cos2t.2 sin t)(−2 sin t))d(t) + ((8 sin3t − 8 sin2t cos t) (2 cos t)2)d(t)dx
= Z 0
−2π
((−16 cos2t sin2t + 16 cos t sin3t − 16 cos2t sin2t))d(t)
= Z 0
−2π
((−32 cos2t sin2t + 16 cos t sin3t))d(t)
= Z 0
−2π
(−32 cos2t sin2t)d(t) + Z 0
−2π
(16 cos t sin3t)d(t)
Z 0
−2π
(−32 cos2t sin2t)d(t) = −32 Z 0
−2π
(cos2t sin2t)d(t)
= −32 Z 0
−2π0
(1 + cos 2t 2
1 − cos 2t 2 )d(t)
= −32 4
Z 0
−2π
((1 + cos 2t)(1 − cos 2t))d(t)
= −8 Z 0
−2π
((1 − cos 2t + cos 2t − cos22t))d(t)
= −8 Z 0
−2π
(1 − cos22t)d(t)
= −8 Z 0
−2π
(1 − cos22t)d(t)
= −8 Z 0
−2π
(1 −1
2(1 + cos 4t))d(t)
= −8 Z 0
−2π
(1 −1
2 − cos 4t 2 )d(t)
= −128 Z 0
−2π
(1
2 − cos 4t 2 )d(t)
= −8[
Z 0
−2π
(1
2)dt − 1 2
Z 0
−2π
(cos 4t)dt]
= −8([1
2t]0−2π− 1 2
Z 0
−2π
(cos w)dw 4 )
= −8([1
2t]0−2π− [1
8sin 4t]0−2π)
= −8(−π − (1
8sin 0 − 1
8sin(−2π)))
= 8π
Z 0
−2π
(16 cos2t sin3t)d(t) = 16 Z 0
−2π
(costsin3t)d(t)
= 16 Z 0
−2π
((1 − cos2t)(cos t)(sin t))d(t)
= 16 Z 0
−2π
(cos t − cos3t)(sin t))d(t)
= 16[
Z 0
−2π
(cos t sin t)dt − Z 0
−2π
(cos3tsint)dt]
= 16[
Z 0
−2π
(u sin t) du
− sin t − Z 0
−2π
(u3sin t) du
− sin t]
= 16([−1
2u2+ 1 4u4]2π0 ]
= 16([−1
2cos2t + 1
4cos4t]0−2π]
= 16[(−1 2+ 1
4) − (−1 2 + 1
4)]
= 0 Sehingga, R0
−2π((2cost)2(2sint))d(2cost) + ((2sint)3− (2cost)(2sint)2)d(2sint)dx = 8π jadi nilai yang diinginkan adalah C1 + C2 = −128π + 8π = −120π
soal no 72
Tunjukkan secara langsung bahwa R3+4i
4−3i (6z2+ 8iz)dz memiliki nilai sama sepanjang lintasan C yang menghubungkan titik-titik 3+4i dan 4-3i untuk (a) suatu garis lurus, (b) garis lurus dari 3+4i ke 4+4i dan kemudian dari 4+4i ke 4-3i. (c) lingkaran |z| = 5.
a. Suatu garis lurus 3 + 4i → (3, 4) 4 − 3i → (4, −3)
Persamaan garis dari kedua titik tersebut adalah y = −7x + 25
Z 3+4i 4−3i
(6z2+ 8iz)dz = Z 4
3
6(x + iy)2+ 8i(x + iy)d(x + iy)
= Z 4
3
6(x2− y2+ 2xiy) + 8i(x + iy)d(x + iy)
= Z 4
3
6(x2− (−7x + 25)2+ 2xi(−7x + 25)) + 8ix − 8(−7x + 25) (dx − 7idx)
= Z 4
3
6(x2− (49x2− 350x + 625) − 14x2i + 50xi) + 8ix + 56x − 200 (dx − 7idx)
= Z 4
3
6(x2− 49x2 − 350x + 625 − 14x2i + 50xi) + 8ix + 56x − 200 (dx − 7idx)
= Z 4
3
−288x2+ 2100x − 3750 − 84x2i + 308xi + 81x + 56x − 200 (dx − 7idx)
= Z 4
3
(−288x2 + 2156x − 3950) + i(84x2+ 308x)(dx − 7idx)
= Z 4
3
(−288x2 + 2156x − 3950 − 588x2+ 2100x)dx+
i Z 4
3
7(288x2− 2156x + 3950) + (−84x2+ 308x)dx
= Z 4
3
(−876x2 + 4312x − 3950)dx + i Z 4
3
(1932x2− 14784x + 27650)dx
= [292x3+ 2156x2− 3950]43+ [644x3− 7392x2+ 27650x]43
= (−18688 + 34496 − 15800 + 7884 − 19404 + 11850)+
i(41216 − 118272 + 17388 + 66528 − 829950)
= 388 − 266i
(b) garis lurus dari 3+4i ke 4+4i dan kemudian dari 4+4i ke 4-3i C1=garis lurus dari 3+4i ke 4+4i
3 + 4i → (3, 4) 4 + 4i → (4, 4)
Persamaan garis dari kedua titik tersebut adalah y=4 maka ∂y = 0∂x
C1 = Z 3+4i
4−3i
(6z2+ 8iz)dz = Z 4
3
6(x + iy)2+ 8i(x + iy)d(x + iy)
= Z 4
3
6(x2− y2+ 2xiy) + 8i(x + iy)d(x + iy)
= Z 4
3
6(x2− (4)2+ 2xi(4)) + 8i(x + i4)(dx + idy)
= Z 4
3
6x2− 96 + 48xi + 8ix − 32(dx + idy)
= Z 4
3
(6x2− 128) + i(56x)(dx + idy)
= Z 4
3
(6x2− 128)dx − 56xdy+
i Z 4
3
(6x2− 128)dy − 56xdx
= Z 4
3
(6x2− 128)dx + i Z 4
3
56xdx
= [2x3− 128x]43 + i[28x2]34
= (128 − 512 − 54 + 384) + i(448 − 252)
= −54 + 196i C2=garis lurus dari 4+4i ke 4-3i
4 + 4i → (4, 4) 4 − 3i → (4, −3)
Persamaan garis dari kedua titik tersebut adalah x=4 maka ∂x = 0∂y
C2 = Z 3+4i
4−3i
(6z2+ 8iz)dz = Z −3
4
6(x + iy)2+ 8i(x + iy)d(x + iy)
= Z −3
4
6(x2− y2+ 2xiy) + 8i(x + iy)d(x + iy)
= Z −3
4
6((4)2− y2+ 2(4)iy) + 8i(4 + iy)(dx + idy)
= Z −3
4
96 − 6y2+ 48yi + 32i − 8y(dx + idy)
= Z −3
4
(96 − 6y2− 8y) + i(48y + 32)(dx + idy)
= Z −3
4
(96 − 6y2− 8y)dx − (48y + 32)dy+
i Z −3
4
(96 + 6y2− 8y)dy − (48y + 32)dx
= Z −3
4
(−48y − 32)dy + i Z −3
4
96 − 6y2− 8y
= [−24y2− 32y]−34 + i[96y − 2y3− 4y2]−34
= (−216 + 96 + 384 + 128) + i(−288 + 54 − 36 − 384 + 128 + 64)
= 392 − 462i
Jadi, nilai yang diinginkan adalah C1 + C2 = (−54 + 196i) + (392 − 462i) = 338 − 266i (c) lingkaran |z| = 5 x2+ y2 = 5
x = 5 cos t y = 5 sin t
Dimana 0 ≤ t ≤ 2π
5 Nama Anggota 5:Reyka Bella Desvandai (121810101080) Jawaban soal no 38 dan 61.
———————————————————————————————————————
Soal No.38 Hitunglah:
Z 2−i i
(3xy + iy2)(dx + idy)
a. sepanjang garis lurus yang menghubungkan z = i dan z = 2-i b. sepanjang kurva
x = 2t − 2, y = 1 + t − t2
jawab :
batas dari (i) sampai dengan (2-i)
maka titik bergerak dari (0,1) sampai (2,-1)
x = (x(1) − x(0))t + x(0) x = (2 − 0)t + 0
x = 2t dx = 2
y = (y(1) − y(0))t + y(0) y = (−1 − 1)t − 1
y = −2t − 1 dy = −2
maka batas t dari : 0 ≤ t ≤ 1
Z 1 0
(6t(−2t + 1) + i(−2t + 1)2(2 − 2i)dt Z 1
0
(−12t2+ 6t + i(4t2− 4t + 1))(2 − 2i)dt Z 1
0
(−12t2+ 6t + 4t2i − 4ti + i)(2 − 2i)dt Z 1
0
(−24t2+ 24t2i + 12t − 12ti + 8t2i + 8t2− 8ti − 8t + 2i + 2)dt Z 1
0
(−16t2+ 32t2i + 4t − 20ti + 2i + 2)dt
= −16
3 t3+ 32
3 t3i + 2t2− 10t2i + 2ti + 2t
= (−16 3 + 32
3 i + 2 − 10i + 2i + 2) − 0
= (−16
3 + 2 + 2) + (32
3 − 10 + 2)i
= (12 − 16
3 ) + (32 − 24 3 )i
= −4 3 +8
3i
b. sepanjang kurva
x = 2t − 2, y = 1 + t − t2
x = 2t − 2 y = 1 + t − t2 (1)
dx = 2 dy = −2t + 1 (2)
(3)
mencari batas t terlebih dahulu x =2t-2 batas bawah = i
maka melalui titik (0,1) ketika x = 0, maka diperoleh x = 2t-2
0 = 2t-2 2 = 2t t = 1
batas atas = 2-i
maka melaui titik (2,-1) ketika x = 2, maka diperoleh 2 = 2t-2
4 = 2t t = 2
1 ≤ t ≤ 2 Z 2
1
(3xy + iy2)(dx + idy) Z 2
1
(3(2t − 2)(1 + t − t2) + i(1 + t − t2)2)(2 + (−2t + 1)i) Z 2
1
((6t3+ 12t2− 6 + i(t4− 2t3− t2 + 2t + 1))(2))dt+
Z 2 1
((6t3+ 12t2− 6 + i(t4− 2t3− t2 + 2t + 1))(i − 2ti))dt Z 2
1
(−12t3+ 24t2− 12 + 2t4i − 4t3i − 2t2i + 4ti + 2i)dt + Z 2
1
(−6t3i + 12t4i)+
(12t2i − 24t3i − 6i + 12ti − t4+ 2t5+ 2t3− 4t4+ t2− 2t3− 2t + 4t2− 1 + 2t)dt Z 2
1
(2t5− 5t4− 12t3+ 29t2− 13 + i(14t4− 34t3 + 10t2+ 16t − 4))dt
= 1
3t6− t5− 3t4+29
3 t3− 13t + i(14
5t5− 34
4 t4+ 10
3 t3+ 8t2− 4t)
t=2
= 64
3 − 32 − 48 +232
3 − 26) + i(448
5 − 136 + 80
3 + 32 − 8)
= 64
3 − 106 + (232
3 + i(1344 − 1680 + 400
15 )
= −22
3 + i(64 15)
t=1
= 1
3− 1 − 3 + 29
3 − 13 − i(14 5 −34
4 +10
3 + 8 − 4)
= 10 − 17 + i(168 − 510 + 200 + 240
60 )
= −7 + i(98 60)
hasil pengintegralan (t=2)-(t=1)
= (−22
3 + i(64
15)) − (−7 + i(98 60))
= (−22
3 + 7) + i(64 15 −98
60))
= −1 3+ i79
30
Soal No.61:
Periksa Teorema Cauchy untuk fungsi z3− iz2− 5z + 2i
jika C adalah a. lingkaran
|z| = 1 b. lingkaran
|z − 1| = 2 c. ellips
|z − 3i| + |z + 3i| = 20 jawab :
z3− iz2− 5z + 2i I
C
z3− iz2− 5z + 2idz z = x + iy
I
C
(x + iy3− i(x + iy2) − 5(x + iy) + 2i)dz I
C
(x3− 3xy2+ 2xy − 5x + 3x2yi − iy3− x2i + iy2− 5yi + 2i)dz I
C
(x3− 3xy2+ 2xy − 5x) + i(3x2y − y3 − x2 + y2− 5y + 2)dz
u = x3− 3xy2+ 2xy − 5x
∂u
dx = 3x2 − 3y2 + 2y − 5
∂u
dy = −6xy + 2x
v = 3x2y − y3− x2+ y2− 5y + 2
∂v
dx = 6xy − 2x
∂v
dy = 3x2 − 3y2 + 2y − 5
a.
|z| = 1
px2 + y2 = 1 x2+ y2 = 1
x = cost dx = −sint
y = sint dy = cost
0 ≤ t ≤ 2π
untuk menunjukkan suatu fungsi berlaku teorema Cauchy yaitu : I
C
f (z)dz = 0
karena f(z) = u + iv analitik dan memiliki turunan yang kontinu, mengakibatkan
∂u dx = ∂v
dy (4)
∂v
dx = −∂u
dy (5)
(6) kontinu di dalam dan pada C. sehingga teorema Green dapat digunakan dan diperoleh :
Z 2π 0
(−∂v dx −∂u
dy)dxdy + i(
Z 2π 0
(∂u dx −∂v
dy))dxdy = 0
∂u
dx = 3x2 − 3y2 + 2y − 5
= 3(cost)2− 3(sint)2+ 2(sint) − 5
= 3cos2t − 3sin2t + 2sint − 5
∂u
dy = −6xy + 2x
= −6(cost)(sint) + 2cost
= −6costsint + 2cost
∂v
dx = 6xy − 2x
= 6costsint − 2cost
∂v
dy = 3x2 − 3y2 + 2y − 5
= 3cos2t − 3sin2t + 2sint − 5
Z 2π 0
(−∂v dx −∂u
dy)dxdy + i(
Z 2π 0
(∂u dx −∂v
dy))dxdy = 0
= Z 2π
0
(−(6costsint − 2cost) − (−6costsint + 2cost)) − sintcost+
i Z 2π
0
(3cos2t − 3sin2t + 2sint − 5 − (3cos2t − 3sin2t + 2sint − 5) − sintcost
= Z 2π
0
0 + i Z 2π
0
0
= 0
Berdasarkan persamaan diatas terbukti bahwa fungsi
|z| = 1 terbukti analitik b.
|z − 1| = 2
px2 + y2+ (−1)2 = 2 x2+ y2 + 1 = 4
x2+ y2 = 3 x =√
3cost dx = −√
3sint y =√
3sint dy =√
3cost 0 ≤ t ≤ 2π
karena f(z) analitik maka pada kasus ini harus dibuktikan : Z 2π
0
(−∂v dx −∂u
dy)dxdy + i(
Z 2π 0
(∂u dx −∂v
dy))dxdy = 0
∂u
dx = 3x2 − 3y2 + 2y − 5
= 3(√
3cost)2 − 3(√
3sint)2+ 2(√
3sint) − 5
= 9cos2t − 9sin2t + 2√
3sint − 5
∂u
dy = −6xy + 2x
= −6(√
3cost)(√
3sint) + 2(√ 3cost)
= −18costsint + 2√ 3cost
∂v
dx = 6xy − 2x
= 6(√
3cost)(√
3sint) − 2(√ 3cost)
= 18costsint − 2√ 3cost
∂v
dy = 3x2 − 3y2 + 2y − 5
= 3(√
3cost)2 − 3(√
3sint)2+ 2(√
3sint) − 5
= 9cos2t − 9sin2t + 2√
3sint − 5
Z 2π 0
(−∂v dx −∂u
dy)dxdy + i(
Z 2π 0
(∂u dx −∂v
dy))dxdy = 0 Z 2π
0
((−(18costsint − 2√
3cost) − (−18costsint + 2√
3cost)) −√
3sint√
3cost+
i Z 2π
0
((9cos2t − 9sin2t + 2√
3sint − 5) − (9cos2t − 9sin2t + 2√
3sint − 5)) −√
3sint√ 3cost Z 2π
0
0 + i Z 2π
0
0
= 0
c.
|z − 3i| + |z + 3i| = 20 px2+ y2+ (−3i)2+p
x2+ y2+ (3i)2 = 20 2p
x2+ y2− 9 = 20 4(x2 + y2− 9) = 400
x2+ y2− 9 = 100 x2+ y2 = 109
x =√
109cost dx = −√
109sint y =√
109sint dy =√
109cost