How do you simplify #6/ (2+3i)#?

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Answer 1 · 2016-01-10T11:04:56

#12/13 - 18/13 i #

To divide this expression we attempt to rationalise the denominator ie. We attempt to make it an integer value. To do this we use the following :

Given the complex number a + bi then the complex conjugate is
a - bi

note the following for multiplying

(a + bi )(a - bi )
#= a^2 +abi - abi - b^2i^2 #
#= a^2 + b^2 ( i =sqrt(-1 )then(( sqrt(-1))^2 = - 1 )#
now #a^2 + b^2 # is a real number

we now multiply the numerator and denominator by (2 - 3i )

# rArr 6/(2 + 3i) . (2 - 3i )/(2 - 3i ) #

#= (6(2 - 3i ))/((2 + 3i)(2 - 3i ))#

multiply out the brackets gives:

#( 12 - 18i)/(4 + 6i -6i - 9i^2) =( 12 - 18i)/( 4+9) =( 12 - 18i)/13#

rewriting in the form a + bi , we get that

#6/(2 + 3i ) = 12/13 - (18i)/13 #