Buenas tardes ncesito resolver stos ejcicios numros complejos:
( 1 - √2i ) + ( - 2 + 3√2i ) =
( 2/5 - 3i ) + ( 7/10 - 3i ) =
( 3 + 4i ) - ( 1 + 3i ) =
( 3/2 + 1/5i ) + ( - 1/3 + 4i ) - ( 1/2 - 1/5i ) =
( 1/4i ) - ( - 5i ) - ( 3i ) =
(3x)+(y-2)i = (5-2x)+(3y-8)i








1

Respuestas

2013-06-04T07:08:14+02:00
(1-√2i)+(-2+3√2i)
=(1-2)+(3√2i-√2i)
=-1+√2i


(2/5-3i)+(7/10-3i)
=(2/5+7/10)+(-3i-3i)
=-3/10-6i


(3+4i)-(1+3i)
=(3 -1)+(4i-3i)
=2 + i


(3/2+1/5i)+(-1/3+4i)-(1/2-1/5i)
=(3/2-1/3-1/2)+(1/5i+4i+1/5i)
=2/3+22/5i

( 1/4i ) - ( - 5i ) - ( 3i ) =9/4i

(3x)+(y-2)i = (5-2x)+(3y-8)i
3x - (5-2x) = - (y-2)i  + (3y-8)i 
3x +2x - 5 = (-y + 2 + 3y - 8)i
5x - 5 = (2y - 6)i