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

1 Answer
Aug 20, 2016

Answer:

#(a+bi)/(c-di)=((ac-bd)/(c^2+d^2))+((ad+bc)/(c^2+d^2))i#

Explanation:

Multiply both numerator and denominator by the complex conjugate of the denominator, then simplify:

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

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

#=(ac+adi+bci+bdi^2)/(c^2-d^2i^2)#

#=(ac+adi+bci-bd)/(c^2+d^2)#

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

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