1 / 9

CLASE 12

CLASE 12. REPRESENTACIÓN GRÁFICA DE LOS NÚMEROS COMPLEJOS . MÓDULO Y ARGUMENTO. I. =3+4 i. z 1. . =3 i. z 4. z 7. z 2. . Afijo. . =2,5. 1. O. z 6. z 5. R. –  5 –  i. =. z 3. . . = – 5+2 i. (3;4). i. . . . . = – 4. = – 2 i. . . . =  a + b.

zeph-jensen
Download Presentation

CLASE 12

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CLASE 12

  2. REPRESENTACIÓN GRÁFICA DE LOS NÚMEROS COMPLEJOS . MÓDULO Y ARGUMENTO .

  3. I =3+4i z1  =3i z4 z7 z2  Afijo  =2,5 1 O z6 z5 R –5–i = z3   =–5+2i (3;4) i     =–4 =–2i . 

  4.  =a +b = 2 2 z=a–bi z CONJUGADOS z (a+bi )(a–bi ) = a2–b2i 2 –1 z y z Tenemos el número complejo tienen igual módulo z=a+bi  : Módulo de z Módulo .

  5. =a +b 2 2 a>0 I b>0 b b a tan= Dado el número complejo 0  z=a+bi Módulo  (a;b) REPRESENTACIÓN GRÁFICA  IC Argumento  a :

  6. z=a+bi  =a +b 2 2 a<0 b>0  (a;b) b a b a  =  tan= z I Módulo   IIC Argumento   0  .

  7. Calcula el módulo y el argumento del número complejo:  =a +b 2 2 I z = –4–3i a= –4 b= –3  2 2 =(–4) +(–3) a<0 b<0 b a =25 tan = =16+9 –3 –4 = z   =180o+   =180o+   = 5 .  = 216,9o III C o  = 5 tan=0,75 =0,75 =36,9o =180o+36,9o = 216,9o

  8. =52 0 5 3 3 4 2 3 ESTUDIO INDIVIDUAL Calcula el módulo y el argumento de los siguientes números complejos: z1= –5+5i z2= –3,7i  =3,7 =135o = 270o i . 3 z4=  1 z3= 2 2  =1  = =300o =0o

  9. . ¿ Módulo y argumento ?  =a +b 2 2 a=23 I b= –2 z = 23–2i  o 2 2 3 =(23 ) +(–2)  z IV C 3 a>0 b<0 b a =16 3 3 3 –2 23  tan = =12+4 =43+4 = = 3  =360o– 360o  =4 tan= =30o  =330o =360o–30o

More Related