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

Question

Christian Antunez · Accepted Answer

#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 #

Serenity Johnson · Answer

undefined To simplify $ \frac{6}{2+3i} $, you can use the process of rationalizing the denominator. First, multiply both the numerator and the denominator by the conjugate of the denominator, which is $ 2-3i $. This yields:

$$ \frac{6}{2+3i} \times \frac{2-3i}{2-3i} = \frac{6(2-3i)}{(2+3i)(2-3i)} $$

Expanding the denominator using the difference of squares formula $ (a+b)(a-b) = a^2 - b^2 $, we get:

$$ (2+3i)(2-3i) = 2^2 - (3i)^2 = 4 - 9i^2 = 4 + 9 = 13 $$

So, the expression simplifies to:

$$ \frac{6(2-3i)}{13} = \frac{12 - 18i}{13} $$

Thus, $ \frac{6}{2+3i} $ simplifies to $ \frac{12 - 18i}{13} $.