How do you divide imaginary numbers?

1 Answer
Apr 25, 2018

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

Explanation:

Suppose we wanted to determine

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

We can multiply the numerator and denominator by the complex conjugate of the denominator. In this case the complex conjugate of the denominator is #c-di#.

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

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

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

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