Pierrot
contestada

Me ayudarian con esto ^^ Suma algebraica de numeros complejos A) (3/2 + 1/5i) + (-1/3 + 4i) - (1/2 - 1/5i)= B) (1/4i) - (-5i) - (3i)= C) (5i) + (-3+4/5i) - (-3/5i) - (-3-4/5i)= D) (-1/3i) + (1/3 + 3i) + (-1/3 - 3i) + (-5i)=

Respuesta :

carlosdario1993

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



(15i+2)/10i + (-1+12i)/3- (5i-2)/10i=

 

(3(15i+2)+10i (-1+12i)-2(5i-2))/30i=

 

(15i+6-10i+120i2-10i+4)/30i=

 

(120i2-5i+4)/30

 

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


1/4i+5i-3i=


1/4i+2i=


(1+8i2)/4i



 

 

 

 

 

 

 

 

 

 

 

 

 

Otras preguntas