How do you write #(a+bi)/(c-di)# in standard form?
1 Answer
Aug 20, 2016
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#
