How do you write #(a+bi)/(c+di)# in standard form?

1 Answer
Nov 29, 2016

Answer:

#(ac+bd)/(c^2+d^2)+(cb-ad)/(c^2+d^2)i#

Explanation:

The standard form is #color(blue)a+color(red)bi#

#(a+bi)/(c+di)#

Multiply the numerator and the denominator by the conjugate of the denominator

#((a+bi)/(c+di))*((c-di)/(c-di))#

#=((a+bi)(c-di))/((c+di)(c-di))#

#=(ac-adi+cbi-bdi^2)/(c^2cancel(-cdi)cancel(+cdi)-d^2i^2)#

Add like terms

#=(ac+(cb-ad)i-bdi^2)/(c^2-d^2i^2)#

#i^2=-1#

#=(ac+(cb-ad)i-bd(-1))/(c^2-d^2(-1))#

#=(ac+(cb-ad)i+bd)/(c^2+d^2)#

#=(ac+bd+(cb-ad)i)/(c^2+d^2)#

#=(ac+bd)/(c^2+d^2)+((cb-ad)i)/(c^2+d^2)#

#=color(blue)((ac+bd)/(c^2+d^2))+color(red)((cb-ad)/(c^2+d^2))i#