Tarea 2

En cada inciso, encontrar el argumento principal de z.

a) z = 1
r = z = √((-1)² + (0)²) = 1
1 = 1 cos θ cos-¹θ = 1/1 = 0
0 = 1 sen θ senθ = 0/1 = 0


b) z = i
r = z = √((0)² + ( 1)²) =
-1 = 1 cos θ cos-¹θ = 90º = π/2
0 = sen θ senθ = 90º = π/2



c) z = - i
r = z = √((o)² + (-1)²) = 1
0 = 1cos θ cos-¹θ = 0/1 = 90º = -π/2
-1 = 1sen θ senθ= -1/1 = 90º = -π/2




d) z = 1 + i
r = z = √((1)² + (1)²) = √2
1 = √ 2 cos θ cos-¹θ = 1/√2 = 45º = π/4
1 = √2 sen θ senθ = 1/√2 = 45º = π/4




e) z= -1 + √3
r = z = √((-1)² + (√3)²) = √4 = 2
-1 = 2 cos θ cos-¹θ= -1/2 = 120º = 2π/3
√3 = 2 sen θ senθ = √3/2 = 120º = 2π/3




f) z = 1 - i
r = z = √(1)² + (-1)² = √2
1 = √2 cos θ cos-¹θ= 1/√2 = 45º = π/4
-1 = √ 2 sen θ senθ = -1/√2 = 45º = -π/4




En cada inciso expresar el numero complejo en forma polar usando su argumento principal.



a) 2i
r = z = √((0)² + (2)²) = √4 = 2
0 = 2cos θ cos- ¹θ = 0/2 = 90º = π/2
2 = 2sen θ senθ = 2/2 = 90º = π/2

2 [ cos (π/2) + i sen (π/2)]



b) -4
r = z = √((-4)² +(0)²)) = √16 = 4
-4 = 4 cos θ cos-¹θ = -4/4 = 180º = π
0 = 4 sen θ senθ = 0/4 = 180º = π

4 [cos π + i sen π]




c) 5 + 5i
r = z = √((5)² + (5)²) = √50
5 = √50 cos θ cos-¹θ = 5/√50 = 45º = π/4
5 = √50 sen θ senθ = 5/√50 = 45º = π/4

√50[cos (π/4) + i sen (π/4) ]



d) -6 + 6√3i
r = z = √((-6)² + (6√3)²) = √38+ 108 = √144 = 12
-6 = 12cos θ cos-¹ θ = -6/12 = 120º = 2π/3
6√3 = 12sen θ sen = 6√3/12 = 120º = 2π/3

12[cos (2π/3) + i sen (2π/3)]



e) -3 -3i
r = z = √((-3)² + (-3)²) = √ 9 + 9 = √18
-3 = √18 cos θ cos-¹θ = -3/√18 = 135º = 3π/4
-3 = √18 sen θ sen θ = -3/√18 = 135º = 3π/4

√18 [cos (3π/4) + i sen (-3π/4)]



f) 2√3 -2i
r = z = √ (2/√3)² + (-2)² = √12 + 4 = √16 = 4
2√ 3 = 4cos θ cos-¹ θ = 2√3/4 = 30º = π/6
-2 = 4 sen θ sen θ = -2/4 = 30º = π/6

4[cos (π/6) + i sen (-π/6)]




Dado que Z1 = 2(cos π/4 + i sen π/4) y Z2 = 3(cos π/6 + i sen π/6) obtener una forma polar de


a) Z1Z2
Z1Z2 =(2)(3)[cos (π/4 + π/6) + i sen (π/4 + π/6)]
= 6[cos (3π/12 + 2π/12) + i sen (3π/12 + 2π/12)]
= 6 ( cos 5π/12 + i sen 5π/12)





b) Z1/Z2
Z1/Z2 = 2/3[cos (π/4 - π/6) + i sen (π/4 - π/6)]
= 2/3[cos 3π/12 - 2π/12) + i sen (3π/12 - 2π/12)]
= 2/3(cos (-π/12) + i sen (-π/12))



c) Z2/Z1
Z2/Z1 = 3/2[cos (π/4 - π/6) + i sen (π/4 + π/6)]
= 3/2[cos (3π/12 - 2π/12) + i sen (3π/12 - 2π/12)]
= 3/2(cos (-π/12) + i sen (-π/12))



d) (Z1)^5/Z2²
(Z1)^5/Z2² = (2)^5/3²[cos (π/4 - π/6) + i sen (π/4 - π/6)]
= 32/9[cos ((5)3π/12) - (2)2π/12) + i sen ((5)3π/12 - (2)2π/12)]
= 32/9[cos (15π/12 - 4π/12) + i sen (15π/12 - 4π/12)]
= 32/9( cos 11π/12 + i sen 11π/12)





En cada inciso, encontrar todas las raices y trazarlas como vectores en el plano complejo


a)√-1
r = z = √(-1)² =
-1 = 1cosθ cos-1θ = -1/1 = -π/2
0 = 1sen θ senθ = 0/1 = - π/2
√1 [cos (π/2)/2 + (2(0)π)/2)) + i sen ((π/2 + (2(0)π)/2]
√1 (cos π/4 + i sen π/4)
√1(-1/√2 + i 1/√2) = 1ra raiz 1/-√2 + 1/√2 i





√1 [cos (π/2)/2 + (2(1)π/2)) + i sen ((π/2 + (2(1)π)/2]
√1( cos π/4 + (2π/2) + i sen (π/4 + 2π/2)
√1(cos 2π/4 + i sen 2π/4)
√1(1/√2 + i -1/√2) = 2da raiz 1/√2 -1/√2 i




b) 1 + √3i
r = z = √(1)² +(√3)² = √1+3 = √ 4 = 2
1 = 2cos θ cos-¹θ= 1/2 = π/3
√3 = 2 senθ senθ = √3/2 = π /3


√2 (cos (π/3)/2 + (2(0)π)/2) + i sen (π/3)/2 + (2(0)π/2))
√2 (cos π/6 + i sen π/6)
√2 (√3/2 + i 1/2) = 1ra raiz √6/2 + √2/2 i


√2(cos (π/6 + (2(1)π)/2) + i sen (π/6 + (2)(1)π)/2)
√2( cos 7π/6) + i sen (7π/6)
√2(-√3/2 - 1/2i) = 2da raiz -√6/2 + √2/2




f) - 8 + 8√3 i
r = z = √(-8)² + (8√3)² = √ 256 = 16
-8 = 16cos θ cos-θ = -8/16 = 120º = 2π/3
8√3 = 16 sen θ senθ = 8√ 3/16 = 120º = 2π/3


4√16(cos (2π/3)/4 + 2(3)π/4) + i sen ((2π/3)/4 + 2(3)π/3)
4√16 (cos 2π/12 + 18π/12) + i sen (2π/12 + 18π/12)
2 ( cos 20π/12 + sen 20π/12)
2(1/2 + i sen (-√3/2) = 1ra raiz 1 + √3i



4√16 (cos (2π/3)/4 + (2(2)π/4)) + i sen (2π/3)/4 + 2(2)π/4)
4√16 (cos 2π/12 + π) + i sen (2π/12 + π)
2( cos 3π/12) + i sen 3π/12)
2(√2/2 + i sen √2/2) = 2da raiz √2 + √2 i



4√16(cos (2π/3)/4 + (2(1)π/4)) + i sen (2π/3)/4 + 2(1)π/4)
4√16( cos 2π/12 + 6π/12) + i sen (2π/12 + 6π/12)
2(cos 8π/12 + i sen 8π/12)
2(-1/2 + i √ 3) = 3ra raiz -1 + √3




4 √16(cos (2π/3)/4 + (2(0)π/4) + i sen (2π/3)/4 + (2(0)π/4)
4√ 16(cos 2π/12 + i sen 2 /12)
2( √3/2 + i 1/2) = 4ta raiz √3 + i