The document discusses complex numbers. It defines the imaginary unit i as the number whose square is -1. It explains that any complex number z can be written in the form z = x + yi, where x is the real part and yi is the imaginary part. It discusses operations like addition, subtraction, conjugation and negation of complex numbers. Graphical representation of complex numbers in the complex plane is also covered.
Se desarrolla el concepto de Inecuaciones a partir de sus principales características; para ello se desarrollan ejemplos que sustentan lo explicado en forma teórica
Se desarrolla el concepto de Inecuaciones a partir de sus principales características; para ello se desarrollan ejemplos que sustentan lo explicado en forma teórica
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email: anasdhyiab@gmail.com
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حلول الاسألة الوزارية
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email: anasdhyiab@gmail.com
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3. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
OBJETIVOS
✓ Extender el campo numérico
✓ Entender la unidad imaginaria
✓ Entender un número complejo y
sus partes.
✓ Operar números complejo.
𝑖2 = −1
Ovillo de Von Koch
4. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
𝑼𝑵𝑰𝑫𝑨𝑫 𝑰𝑴𝑨𝑮𝑰𝑵𝑨𝑹𝑰𝑨
La unidad imaginaria (i), denotado por Euler en
1777, se define como el número cuyo cuadrado
resulta -1.
Potencias enteras de i
𝑖2
= −1
𝑖 = −1
NOTA Para facilitar el aprendizaje
Se definen: 𝑖0
= 1 𝑖1
= 𝑖
𝑖1
= 𝑖
𝑖2
= −1
𝑖3
= −𝑖
𝑖4 = 1
𝑖5 = 𝑖
𝑖6
= −1
𝑖7
= −𝑖
𝑖8
= 1
𝑖9 = 𝑖
𝑖10 = −1
𝑖11 = −𝑖
𝑖12
= 1
luego
Se deduce que:
1) 𝑖4𝑛
= 1; 𝑛 ∈ ℤ
2) 𝑖4𝑛+𝑟
= 𝑖𝑟
; 𝑟; 𝑛 ⊂ ℤ
3) 𝑖𝑛
+ 𝑖𝑛+1
+ 𝑖𝑛+2
+ 𝑖𝑛+3
= 0; 𝑛 ∈ ℤ
Ejemplo
𝑅𝑒𝑑𝑢𝑐𝑖𝑟: 𝑀 = 𝑖 + 𝑖2
+ 𝑖3
+ 𝑖4
+ 𝑖5
𝑀 = 𝑖 + 𝑖2 + 𝑖3 + 𝑖4 + 𝑖5
0 𝑖
= 𝑖
(𝑖)
NOTA
𝑖−1
𝑖−1
= −𝑖
=
1
𝑖
×
𝑖
𝑖
=
𝑖
−1
= −𝑖
5. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Resultados importantes
1) 1 + 𝑖 2
= 12
+ 2 1 𝑖 + 𝑖2
−1
1 + 𝑖 2 = 1 + 2𝑖 − 1 = 2𝑖
1 + 𝑖 2
= 2𝑖
2) 1 − 𝑖 2 = 12 − 2 1 𝑖 + 𝑖2
−1
1 − 𝑖 2 = 1 − 2𝑖 − 1 = −2𝑖
1 − 𝑖 2 = −2𝑖
3)
1 + 𝑖
1 − 𝑖
(−1)
4)
1 − 𝑖
1 + 𝑖
1 + 𝑖
1 − 𝑖
= 𝑖
1 − 𝑖
1 + 𝑖
= −𝑖
=
1 + 𝑖
1 − 𝑖
×
1 + 𝑖
1 + 𝑖
=
1 + 𝑖 2
12 − 𝑖2 =
2𝑖
2
= 𝑖
(−1)
=
1 − 𝑖
1 + 𝑖
×
1 − 𝑖
1 − 𝑖
=
1 − 𝑖 2
12 − 𝑖2 =
−2𝑖
2
= −𝑖
5) 𝑏 − 𝑎𝑖 𝑖 = 𝑏𝑖 − 𝑎 𝑖2
(−1)
𝑏 − 𝑎𝑖 𝑖 = 𝑏𝑖 + 𝑎
𝑖 =
𝑎 + 𝑏𝑖
𝑏 − 𝑎𝑖
Luego:
5 + 7𝑖
7 − 5𝑖
= 𝑖
6. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
𝑭𝑶𝑹𝑴𝑨 𝑩𝑰𝑵Ó𝑴𝑰𝑪𝑨
Todo número complejo z tiene la forma
𝑧 = 𝑥 + 𝑦𝑖
𝑑𝑜𝑛𝑑𝑒:
Es decir:
𝑅𝑒 𝑧 = 𝑥 ∶ 𝑃𝑎𝑟𝑡𝑒 𝑟𝑒𝑎𝑙
I𝑚 𝑧 = 𝑦 ∶ 𝑃𝑎𝑟𝑡𝑒 𝑖𝑚𝑎𝑔𝑖𝑛𝑎𝑟𝑖𝑎
𝑖2
= −1
𝑥; 𝑦 ∈ ℝ ;
Nota
El conjunto de los números complejos; se denota por:
ℂ = 𝑥 + 𝑦𝑖 ∶ 𝑥𝜖ℝ ∧ 𝑦𝜖ℝ
Ejemplos Tenemos:
1) 𝑧1 = 3 − 7𝑖
𝑅𝑒 𝑧1 = 3 𝐼𝑚 𝑧1 = −7
2) 𝑧2 =
3
2
+ 5𝑖
𝑅𝑒 𝑧2 =
3
2
𝐼𝑚 𝑧2 = 5
3) 𝑧2 = 0 − 9𝑖
𝑅𝑒 𝑧2 = 0 𝐼𝑚 𝑧2 = −9
4) 𝑧2 = −4 + 0𝑖
𝑅𝑒 𝑧2 = −4 𝐼𝑚 𝑧2 = 0
7. C R E E M O S E N L A E X I G E N C I A
𝑹𝑬𝑷𝑹𝑬𝑺𝑬𝑵𝑻𝑨𝑪𝑰Ó𝑵 𝑮𝑹Á𝑭𝑰𝑪𝑨
Los números complejos se pueden ubicar en el plano
complejo, que está compuesto por el eje real y el eje
imaginario.
𝑧 = 𝑥 + 𝑦𝑖
𝐸𝑗𝑒 𝑅𝑒𝑎𝑙
𝐸𝑗𝑒 𝑖𝑚𝑎𝑔𝑖𝑛𝑎𝑟𝑖𝑜
𝑥
𝑦
Ejemplos
𝑅𝑒(𝑧)
𝐼𝑚(𝑧)
3
5
−2
−3
2
5
−3
−4
𝑧1 = 3 + 2𝑖
𝑧2 = −3 + 5𝑖
𝑧3 = −2 − 4𝑖
𝑧4 = 5 − 3𝑖
Grafique
𝑧1 = 3 + 2𝑖 ; 𝑧2 = −3 + 5𝑖 ; 𝑧3 = −2 − 4𝑖 ; 𝑧4 = 5 − 3𝑖
𝐷𝑎𝑑𝑜: 𝑧 = 𝑥 + 𝑦𝑖, 𝑠𝑢 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑎𝑐𝑖ó𝑛 𝑔𝑟á𝑓𝑖𝑐𝑎 𝑒𝑠
8. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Tipos de números complejos Diremos que 𝑧 = 𝑥 + 𝑦𝑖, es
1) 𝐶𝑜𝑚𝑝𝑙𝑒𝑗𝑜 𝑟𝑒𝑎𝑙 2) 𝐶𝑜𝑚𝑝𝑙𝑒𝑗𝑜 𝑖𝑚𝑎𝑔𝑖𝑛𝑎𝑟𝑖𝑜 𝑝𝑢𝑟𝑜
Ejemplos
𝑎) 𝑧 = 6 + 0𝑖 = 6
𝑏) 𝑤 = −7 + 0𝑖 = −7
3) 𝐶𝑜𝑚𝑝𝑙𝑒𝑗𝑜 𝑛𝑢𝑙𝑜 𝑠𝑖 𝑥 = 0 ∧ 𝑦 = 0
Ejemplos
𝑅𝑒
𝐼𝑚
6
−7
𝑎) 𝑧 = 0 + 3𝑖 = 3𝑖
𝑏) 𝑤 = 0 − 4𝑖 = −4𝑖
𝑧 = 6 + 0𝑖
𝑤 = −7 + 0𝑖
𝑅𝑒
𝐼𝑚
𝑅𝑒
𝐼𝑚
𝑧 = 0 + 0𝑖
𝑧 = 0 + 0𝑖
−4
3 𝑧 = 0 + 3𝑖
𝑤 = 0 − 4𝑖
𝑠𝑖 𝑥 ≠ 0 ∧ 𝑦 = 0 𝑠𝑖 𝑥 = 0 ∧ 𝑦 ≠ 0
9. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Tipos de números complejos
𝑆𝑖 𝑧 = 𝑥 + 𝑦𝑖
Complejo conjugado ( ҧ
𝑧)
ҧ
𝑧 = 𝑥 − 𝑦𝑖 𝑆𝑖 𝑧 = 𝑥 + 𝑦𝑖
Complejo opuesto (𝑧∗)
𝑧∗
= −𝑥 − 𝑦𝑖
Ejemplos Ejemplos
1) 𝑧1 = 3 − 7𝑖 ഥ
𝑧1 = 3 + 7𝑖
2) 𝑧2 = −4 + 5𝑖 ഥ
𝑧2 = −4 − 5𝑖
3) 𝑧3 = 0 − 6𝑖 ഥ
𝑧3 = 0 + 6𝑖
4) 𝑧4 = 6 + 0𝑖 ഥ
𝑧4 = 6 − 0𝑖
𝑧3 = −6𝑖 ഥ
𝑧3 = 6𝑖
𝑧4 = 6 ഥ
𝑧4 = 6
1) 𝑧1 = 4 − 5𝑖 𝑧1
∗ = −4 + 5𝑖
2) 𝑧2 = −3 + 7𝑖 𝑧2
∗
= 3 − 7𝑖
3) 𝑧3 = 0 − 5𝑖 𝑧3
∗
= −0 + 5𝑖
𝑧3 = −5𝑖 𝑧3
∗ = 5𝑖
4) 𝑧4 = 8 + 0𝑖 𝑧4
∗
= −8 − 0𝑖
𝑧4 = −5𝑖 𝑧4
∗
= −8 − 0𝑖
NOTA: 𝑧 = ҧ
𝑧 𝑧 𝑒𝑠 𝑐𝑜𝑚𝑝𝑙𝑒𝑗𝑜 𝑟𝑒𝑎𝑙
10. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Operaciones en ℂ
Luego:
I) Igualdad de complejos:
𝑧1 = 𝑧2
𝑧1 = 𝑎 + 𝑏𝑖 𝑎; 𝑏; 𝑚; 𝑛 ∈ ℝ
𝑧2 = 𝑚 + 𝑛𝑖
𝑎 = 𝑚 𝑏 = 𝑛
∧
Ejemplo
∧ 𝑧2 = 𝑎 + 𝑏 + 𝑎𝑏𝑖
𝑆𝑖: 𝑧1 = 𝑧2.
Resolución
𝑧1= 5 + 3𝑖
𝐷𝑎𝑑𝑜
𝐶𝑎𝑙𝑐𝑢𝑙𝑒 𝑎2 + 𝑏2
Tenemos que: 𝑧1 = 𝑧2
5 + 3𝑖 = 𝑎 + 𝑏 + 𝑎𝑏𝑖
5 = 𝑎 + 𝑏 ∧ 3 = 𝑎𝑏
Sabemos
(𝑎 + 𝑏)2
= 𝑎2
+ 𝑏2
+ 2𝑎𝑏
5 3
52
= 𝑎2
+ 𝑏2
+ 2(3) 25 = 𝑎2
+ 𝑏2
+ 6
𝑎2
+ 𝑏2
= 19
∴
; 𝑎; 𝑏 ∈ ℝ
11. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
𝑧1 + 𝑧2 = (𝑎 + 𝑚) + (𝑏 + 𝑛)𝑖
II) Adición:
Ejemplo
𝑧1 = 5 + 7𝑖
𝑧2 = −4 + 3𝑖
𝑧1 + 𝑧2 = 1 + 10𝑖
(+)
𝑧1 = 5 + 7𝑖 ; 𝑧2 = −4 + 3𝑖
Sean
𝑧1 − 𝑧2 = (𝑎 − 𝑚) + (𝑏 − 𝑛)𝑖
III) Sustracción:
Ejemplo
𝑧1 = 3 − 4𝑖
𝑧2 = 8 + 9𝑖
𝑧1 − 𝑧2 = −5 − 13𝑖
(−)
𝑧1 = 3 − 4𝑖 𝑧2 = 8 + 9𝑖
𝑧1 = 𝑎 + 𝑏𝑖 𝑎; 𝑏; 𝑚; 𝑛 ∈ ℝ
𝑧2 = 𝑚 + 𝑛𝑖 𝑧1 = 𝑎 + 𝑏𝑖 𝑎; 𝑏; 𝑚; 𝑛 ∈ ℝ
𝑧2 = 𝑚 + 𝑛𝑖
Sean
12. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
𝑧1. 𝑧2 = (𝑎𝑚 − 𝑏𝑛) + (𝑎𝑛 + 𝑏𝑚)𝑖
IV) Multiplicación: Ejemplo
𝑧1. 𝑧2 = 2 + 3𝑖 4 + 5𝑖
𝑧1 = 2 + 3𝑖 ; 𝑧2 = 4 + 5𝑖
𝑧1. 𝑧2 = 8 + 10𝑖 + 12𝑖 − 15
𝑧1. 𝑧2 = −7 + 22𝑖
𝑧1 = 𝑎 + 𝑏𝑖 𝑎; 𝑏; 𝑚; 𝑛 ∈ ℝ
𝑧2 = 𝑚 + 𝑛𝑖
Luego
𝑧1. 𝑧2 = 𝑎 + 𝑏𝑖 𝑚 + 𝑛𝑖
𝑧1. 𝑧2 = 𝑎𝑏 + 𝑎𝑛𝑖 + 𝑏𝑚𝑖 + 𝑏𝑛 𝑖2
−1
𝑧1. 𝑧2 = 𝑎𝑏 + 𝑎𝑛𝑖 + 𝑏𝑚𝑖 − 𝑏𝑛
Sean
𝐸𝑛𝑐𝑢𝑒𝑛𝑡𝑟𝑒 𝑒𝑙 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑧1. 𝑧2
Resolución
𝑧1. 𝑧2 = 2.4 + 2.5𝑖 + 3𝑖. 4 + 3.5𝑖2
−1
13. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
V) División:
𝑧1
𝑧2
=
𝑎 + 𝑏𝑖
𝑚 + 𝑛𝑖
=
𝑎 + 𝑏𝑖
𝑚 + 𝑛𝑖
×
𝑚 − 𝑛𝑖
𝑚 − 𝑛𝑖
=
𝑎𝑚 + 𝑏𝑛 + 𝑏𝑚 − 𝑎𝑛 𝑖
𝑚2 + 𝑛2
=
𝑎𝑚 + 𝑏𝑛
𝑚2 + 𝑛2
+
𝑏𝑚 − 𝑎𝑛
𝑚2 + 𝑛2
𝑖
𝑧1
𝑧2
𝑧1
𝑧2
Ejemplo
𝑧1 = 3 + 2𝑖 ; 𝑧2 = 4 + 3𝑖
Resolución
𝑧1
𝑧2
=
3 + 2𝑖
4 + 3𝑖
×
4 − 3𝑖
4 − 3𝑖
=
3 + 2𝑖 4 − 3𝑖
4 + 3𝑖 4 − 3𝑖
=
𝑎𝑚 + 𝑏𝑛 + 𝑏𝑚 − 𝑎𝑛 𝑖
𝑚2 − 𝑛2𝑖2
𝑧1
𝑧2
−1
𝑧1
𝑧2
=
12 − 9𝑖 + 8𝑖 + 6
42 + 32
𝑧1
𝑧2
=
18 − 𝑖
42 + 32
25
𝑧1
𝑧2
=
18
25
−
𝑖
25
14. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
TEOREMA
𝑎 + 𝑏𝑖
𝑚 + 𝑛𝑖
𝑆𝑖 𝑒𝑠 𝑢𝑛 𝑐𝑜𝑚𝑝𝑙𝑒𝑗𝑜 𝑟𝑒𝑎𝑙.
𝑎
𝑚
=
𝑏
𝑛
DEMOSTRACIÓN
𝑇𝑒𝑛𝑒𝑚𝑜𝑠 𝑞𝑢𝑒
𝑎 + 𝑏𝑖
𝑚 + 𝑛𝑖
= 𝑘 ; 𝑘 ∈ ℝ
𝑂𝑝𝑒𝑟𝑎𝑛𝑑𝑜
𝑎 + 𝑏𝑖 = 𝑘 𝑚 + 𝑛𝑖
𝑎 + 𝑏𝑖 = 𝑘𝑚 + 𝑘𝑛𝑖
𝑎 = 𝑘𝑚 ∧ 𝑏 = 𝑘𝑛
𝑘 =
𝑎
𝑚
𝐿𝑢𝑒𝑔𝑜
𝑘 =
𝑎
𝑚
=
𝑏
𝑛
𝐸𝑛𝑡𝑜𝑛𝑐𝑒𝑠
𝑘 =
𝑏
𝑛
∧
Ejemplo
𝑎; 𝑏; 𝑚; 𝑛 ∈ ℝ
NOTA:
𝑎 + 4𝑖
3 + 𝑏𝑖
𝑆𝑖 𝑒𝑠 𝑢𝑛 𝑐𝑜𝑚𝑝𝑙𝑒𝑗𝑜 𝑟𝑒𝑎𝑙.
𝑑𝑜𝑛𝑑𝑒 𝑎 𝑦 𝑏 𝑠𝑜𝑛 𝑛ú𝑚𝑒𝑟𝑜𝑠 𝑟𝑒𝑎𝑙𝑒𝑠.
𝐸𝑛𝑐𝑢𝑒𝑛𝑡𝑟𝑒 𝑒𝑙 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑎. 𝑏
Resolución
𝑃𝑜𝑟 𝑡𝑒𝑜𝑟𝑒𝑚𝑎
𝑎
3
=
4
𝑏
𝑎. 𝑏 = 12
15. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
TEOREMA
𝑎 + 𝑏𝑖
𝑚 + 𝑛𝑖
𝑆𝑖 𝑒𝑠 𝑢𝑛 𝑐𝑜𝑚𝑝𝑙𝑒𝑗𝑜
−
𝑎
𝑛
=
𝑏
𝑚
DEMOSTRACIÓN
𝑇𝑒𝑛𝑒𝑚𝑜𝑠 𝑞𝑢𝑒
𝑎 + 𝑏𝑖
𝑚 + 𝑛𝑖
= 𝑘𝑖 ; 𝑘 ∈ ℝ
𝑂𝑝𝑒𝑟𝑎𝑛𝑑𝑜
𝑎 + 𝑏𝑖 = 𝑘𝑖 𝑚 + 𝑛𝑖
𝑎 + 𝑏𝑖 = 𝑘𝑚𝑖 − 𝑘𝑛
𝑎 = −𝑘𝑛 ∧ 𝑏 = 𝑘m
𝑘 = −
𝑎
𝑛
𝐿𝑢𝑒𝑔𝑜
𝑘 = −
𝑎
𝑛
=
𝑏
𝑚
𝐸𝑛𝑡𝑜𝑛𝑐𝑒𝑠
𝑘 =
𝑏
𝑚
∧
Ejemplo
𝑖𝑚𝑎𝑔𝑖𝑛𝑎𝑟𝑖𝑜 𝑝𝑢𝑟𝑜.
𝑎; 𝑏; 𝑚; 𝑛 ∈ ℝ
NOTA:
𝑎 + 𝑏𝑖
−3 + 4𝑖
𝑆𝑖 𝑒𝑠 𝑢𝑛 𝑐𝑜𝑚𝑝𝑙𝑒𝑗𝑜
𝑖𝑚𝑎𝑔𝑖𝑛𝑎𝑟𝑖𝑜 𝑝𝑢𝑟𝑜.
𝐷𝑜𝑛𝑑𝑒 𝑎 𝑦 𝑏 𝑠𝑜𝑛 𝑛ú𝑚𝑒𝑟𝑜𝑠 𝑟𝑒𝑎𝑙𝑒𝑠.
𝐸𝑛𝑐𝑢𝑒𝑛𝑡𝑟𝑒 𝑒𝑙 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒
𝑎
𝑏
Resolución
𝑃𝑜𝑟 𝑡𝑒𝑜𝑟𝑒𝑚𝑎
−
𝑎
−3
=
𝑏
4
𝑎
𝑏
=
3
4
16. w w w . a c a d e m i a c e s a r v a l l e j o . e d u . p e
17. w w w . a c a d e m i a c e s a r v a l l e j o . e d u . p e