math.com
Home    |    Teacher    |    Parents    |    Glossary    |    About Us
Homework Help Practice Ask An Expert Calculators & Tools Games Store
Email this page to a friend Email this page to a friend
Resources
· Cool Tools
· Formulas & Tables
· References
· Test Preparation
· Study Tips
· Wonders of Math
 
Search


  
Complejidad
(Matemática | Cabos sueltos | Complejidad)

Números complejos

Operaciones básicas

i = sqrt(-1)

i^2 = -1

1 / i = -i

i^(4k) = 1; i^(4k+1) = i; i^(4k+2) = -1; i^(4k+3) = -i (k = un entero)

sqrt( i ) = sqrt(1/2)+ sqrt(1/2) i


Definiciones de funciones y operaciones complejas

(a + bi) + (c + di) = (a+c) + (b + d) i

(a + bi) (c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad +bc) i

1/(a + bi) = a/(a^2 + b^2) - b/(a^2 + b^2) i

(a + bi) / (c + di) = (ac + bd)/(c^2 + d^2) + (bc - ad)/(c^2 +d^2) i

a2 + b2 = (a + bi) (a - bi)   (la suma de cuadrados)

e^(i theta) = costheta + i sen theta

n^(a + bi) = (cos(b ln n) + i sen(b ln n))n^a

si z = r(cos theta+ i sen theta) pues z^n = r^n ( cos ntheta+ i sen ntheta )(El Teorema de DeMoivre)

si w = r(cos theta+ i sen theta); n=un entero. entonces hay n, nth raices complejas (z) de w para k=0,1,..n-1:

z(k) = r^(1/n) [ cos( (theta+ 2(PI)k)/n ) + i sen( (theta+ 2(PI)k)/n ) ]

si z = r (cos theta+ i sen theta) pues ln(z) = ln r + i theta

sen(a + bi) = sen(a)cosh(b) + cos(a)senh(b) i

cos(a + bi) = cos(a)cosh(b) - sen(a)senh(b) i

tan(a + bi) = ( tan(a) + i tanh(b) ) / ( 1 - i tan(a) tanh(b)) = ( sech^2(b)tan(a) + sec^2(a)tanh(b) i ) / (1 + tan^2(a)tanh^2(b))

  
 
  

 
Contact us | Advertising & Sponsorship | Partnership | Link to us

© 2000-2023 Math.com. All rights reserved.     Legal Notices.     Please read our Privacy Policy.