CAPITULO XXXII
CANTIDADES IMAGINARIAS
CANTIDADES IMAGINARIAS
- Ejercicio 263
Dividir:
- ( 1+ β1 ) Γ· ( 1β β1 ) 1+ β1 1β β1 = 1+ β1 1β β1 Γ 1+ β1 1+ β1 = ( 1+ β1 ) 2 1 2 β ( β1 ) 2 = 1+2 β1 + ( β1 ) 2 1β( β1 ) = 1 +2 β1 β 1 1+1 = 2 β1 2 =i
- ( 3+ β1 ) Γ· ( 3β β1 ) = 3+ β1 3β β1 Γ 3+ β1 3+ β1 = ( 3+ β1 ) 2 3 2 β ( β1 ) 2 = 9+6 β1 + ( β1 ) 2 9β( β1 ) = 9+6 β1 β1 9+1 = 8+6 β1 10 = 2 ( 4+3i ) = 4+3i 5
- ( 5β3 β1 ) Γ· ( 3+4 β1 ) = 5β3 β1 3+4 β1 Γ 3β4 β1 3β4 β1 = ( 5β3i ) ( 3β4i ) 3 2 β ( 4i ) 2 = 15β20iβ9i+12 i 2 9β16 i 2 = 15β29i+12( β1 ) 9β16( β1 ) = 15β29iβ12 9+16 = 3β29i 25
- ( 8β5i ) Γ· ( 7+6i ) = 8β5i 7+6i Γ 7β6i 7β6i = ( 8β5i ) ( 7β6i ) 7 2 β ( 6i ) 2 = 56β48iβ35i+30 i 2 49β36 i 2 = 56β83i+30( β1 ) 49β36( β1 ) = 56β83iβ30 49+36 = 26β83i 85
- ( 4+ β3 ) Γ· ( 5β4 β3 ) = 4+ β3 5β4 β3 Γ 5+4 β3 5+4 β3 = ( 4+ 3 i ) ( 5+4 3 i ) 5 2 β ( 4 β3 ) 2 = 20+16 3 i+5 3 i+4( 3 i 2 ) 25β16( β3 ) = 20+21 3 i+12 i 2 25+48 = 20+21 3 i+12( β1 ) 73 = 20+21 3 iβ12 73 = 8+21 3 i 73
- ( 2 +2 β5 ) Γ· ( 4 2 β β5 ) = 2 +2 β5 4 2 β β5 Γ 4 2 + β5 4 2 + β5 = 2 +2 5 i 4 2 β 5 i Γ 4 2 + 5 i 4 2 + 5 i = ( 2 +2 5 i ) ( 4 2 + 5 i ) ( 4 2 ) 2 β ( 5 i ) 2 = 4( 2 ) + 10 i+8 10 i+2( 5 ) i 2 16( 2 ) β5 i 2 = 8+9 10 i+10( β1 ) 32β5( β1 ) = 8+9 10 iβ10 32+5 = β2+9 10 i 37