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

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Answer 1 · 2016-09-21T16:14:58

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

To write #(a-bi)/(c-di)# in standard form, what is needed is first rationalize the denominator.

To rationalize denominator, we need to multiply it by the conjugate complex of denominator #c-di#, which is #c+di#. Hence,

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

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

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

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

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

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