UNIGA matematika

UTS matematika
1.Z1 = 2+2i√3 dan z2 √ - 3
 mod ( (z1 )/z2) = r1cisθ1/r1cisθ2
∴ r1 = √((2)^2+(2√3)^2 )
∴ r2 = √(4+12)
= √16 = 4
 θ1= tan^(-1)⁡〖(2√3)/2〗
= 45° = 1/4 π
θ2 = tan^(-1)⁡〖(-1)/√3〗
= 30° = - 1/6 π
∴z1/z2 = r1cisθ1/v2cisθ2 = (4cis□(1/4) π)/(2cis-□(1/6) π) = □(4/2)cis (θ1-θ2)
=2 cis (□(1/4) π-(-□(1/6) π)
=2 cis (□(1/4) π+□(1/6) π)
=2 cis □(5/12 π)

2.z3 = -2+2i⟶ z¬¬3ω⟶ z=ω□(1/3)
Z = (-2+2i) "1" /"3"
r = √((-2)^2+(2)^2 )
√(4+4) = √8 = 2√3
θ = tan^(-1)⁡〖y/x〗
= 2/(-2) = -1 = - 450 = = -1/4 π

Zo = √(n&r)(cos⁡〖(□((θ+2kπ)/n))+i sin ((θ+2kπ)/n)〗 )
∛r (cos⁡〖(□((θ+2kπ)/3))+i sin ((θ+2kπ)/3)〗 )
∛(2√3) (cos⁡〖( □((-□(1/4)+2kπ)/3))+i sin ((-□(1/4) π+2.0.π)/3)〗 )
2√(6&3)((cos-15π)⁡〖+ i sin (- 1/12 π)〗 )
2√(6&3).(0,96⁡〖+ i .-0,26〗 )
1,92 √(6&3) - i.0,52

Zi = √(n&r)(cos⁡〖(□((θ+2kπ)/n))+i sin ((θ+2kπ)/n)〗 )
∛(2√3) (cos⁡〖( □((-□(1/4)+2.1.π)/3))+i sin ((-□(1/4) π+2.1.π)/3)〗 )
2√(6&3)((cos-105°)⁡〖+ i sin (105°)〗 )
2√(6&3)(-0,25) + i (0,95)
-0,5 √(6&3) - 1,92 i

Zo = √(n&r)(cos⁡〖(□((θ+2kπ)/n))+i sin ((θ+2kπ)/n)〗 )
∛r (cos⁡〖(□((θ+2kπ)/3))+i sin ((θ+2kπ)/3)〗 )
∛(2√3) (cos⁡〖( □((-□(1/4)+2kπ)/3))+i sin ((-□(1/4) π+2.0.π)/3)〗 )
2√(6&3)((cos-15π)⁡〖+ i sin (- 1/12 π)〗 )
2√(6&3).(0,96⁡〖+ i .-0,26〗 )
1,92 √(6&3) - i.0,52

Z2 = √(n&r)(cos⁡〖(□((θ+2kπ)/n))+i sin ((θ+2kπ)/n)〗 )
∛(2√3) (cos⁡〖( □((-□(15/4) π)/3))+i sin ((-□(15/4) π)/3)〗 )
∛(2√3) (cos⁡〖( □((-15)/12))+i sin ((-15)/12)〗 )
- 0,5 cos⁡〖(-225)+i sin⁡〖(-225)〗 〗
-0,5(-0,71 + i 0,71)
0,355 + ( - i 0,355)
0,355 - i 0,355

3. f(z) = (x3- 3xy2-x2+y2)+ i (3x2y-y3-2xy)
a. ⋃_(x,y)▒= x3- 3xy2-x2+y2 → ⋃_(x,y)▒= 3x2y-y3-2xy
∂μ/∂x = (∂⋁)/∂γ dan (∂⋁)/∂x = - ∂μ/∂γ
∂μ/∂x = 3x2-3y-2x2 → (∂⋁)/∂x = 6xy – 2y
(∂⋁)/∂γ = 3x2-3y2-2x → (∂⋁)/∂x = 3x2-3y2-2x
∂μ/∂x= ∂μ/∂γ = 3x2-3y2-2x
b. (∂⋁)/∂γ = - (∂⋁)/∂x =beda ∴ fungsi tidak analitik

4.e^z = 3+3i√3
Misalkan : z = μ + iy = e^z = e^μ (cos⁡〖y+i sin⁡y 〗 )..
Misalkan : w = 3+3√3
r = √((3)^2+(3√3)^2 ) = √(9+27) = √36 = 6
θ= tan^(-1)⁡〖(3√3)/3〗 = 600 = 1/3 π
W = r(cos θ + i sin θ)
W = 6(cos 1/3 π+ i sin 1/3 π)…
Dari  dan  didapat
e^z = e^μ (cos⁡〖y+i sin⁡y 〗 )= 6(cos 1/3 π+ i sin 1/3 π)
e^μ=6 → μ=Ln 6
y = 1/3 π → y= 1/3 π+2kπ
∴ z = Ln 6+i(1/3 π+2kπ)

5.a.cos(2-2i√3) =
(e^(i(2+2i√3))+e^(-i(2+2i√3)))/2
(e^(i(2i+2√3))+e^(-i(2i+2√3)))/2
(e^(2√3).e^2i+e^(2√3).e^(- 2i))/2
(12,8(cos⁡〖2+i sin2〗 )-12,8(cos⁡〖2-i sin2〗 ) )/2
(12,8(0,99⁡〖+i 0,03〗 )-12,8(0,99⁡〖-i 0,03〗 ) )/2
(1 2,67+0,38 i- 12,67+0,38 i)/2 = 0/2 = 02
b.Ln (√3+ i )=