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

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Answer 1 · 2015-11-30T10:10:58

#2/3 - i#

You must use something that resembles rationalization with roots at the denominator: multiply both numerator and denominator by #3-3i#:

#(5-i)/(3+3i) * (3-3i)/(3-3i) = ((5-i)(3-3i))/((3+3i)(3-3i))#

and use the fact that

#(3+3i)(3-3i) = 3^2 - (3i)^2 = 9-9i^2=9+9=18#

Then, expand the numerator as usual:

#(5-i)(3-3i)#

# = 15-15i-3i+3i^2#

# = 15-18i-3 #

#= 12-18i#

We can simplify something:

#(12-18i)/18 = 2/3 - i#