How do you simplify #(4+5i)/(2-3i)#?

1 Answer
sente
Feb 13, 2016

#(4+5i)/(2-3i) =-7/13+22/13i#

Explanation:

Given a complex number #z = a+bi#, the complex conjugate of #z#, denoted #bar(z)#, is #bar(z) = a-bi#. For any complex number, #zbar(z)# is a real number. Thus, we can eliminate the complex number from the denominator by multiplying the numerator and the denominator by the conjugate of the denominator.

#(4+5i)/(2-3i) = ((4+5i)(2+3i))/((2-3i)(2+3i))#

#=(-7+22i)/13#

#=-7/13+22/13i#

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