How do you simplify #\frac { 2+ 8i } { 4- 3i }#?

1 Answer
Sridhar V.
Nov 28, 2017

#-(16/25)+(38/25)i#

Explanation:

While working on complex numbers, we must remember the following useful properties:

#i = sqrt(-1)#

#i^2 = (-1)#

#i^3 = -sqrt(-1)#

#i^4 = (+1)#

If we want to divide a complex number by another complex number, then we must multiply both the numerator and the denominator by the "Complex Conjugate" of the denominator.

"Complex Conjugate" of a complex number with equal real part and imaginary part equal in magnitude but opposite in sign.

This process will eliminate the imaginary part in the denominator, as we can see below:

#(2 + 8i)/(4-3i)X(4+3i)/(4+3i)#

#= ((2 + 8i)(4+3i) )/ (( 4-3i)(4+3i))# {Distribute( use FOIL ) to remove parentheses}

#= (8 + 6i + 32i + 24(-1))/(4^2 - (3i)^2)#

#= (8 + 38i - 24) / (16 -9(-1))#

#= (38i - 16)/25#

Next step is to write our expression in #(a + bi)# form.

#-(16/25) + (38/25)i#

Now we have our answer.

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