Question 297082

{{{(2+6i)/(3-3i)}}} Start with the given expression.



{{{((2+6i)/(3-3i))((3+3i)/(3+3i))}}} Multiply the fraction by {{{(3+3i)/(3+3i)}}}.



{{{((2+6i)(3+3i))/((3-3i)(3+3i))}}} Combine the fractions.



{{{((2)(3)+(2)(3i)+(6i)(3)+(6i)(3i))/((3-3i)(3+3i))}}} FOIL the numerator.



{{{((2)(3)+(2)(3i)+(6i)(3)+(6i)(3i))/((3)(3)+(3)(3i)+(-3i)(3)+(-3i)(3i))}}} FOIL the denominator.



{{{(6+6i+18i+18i^2)/(9+9i-9i-9i^2)}}} Multiply.



{{{(6+6i+18i+18(-1))/(9+9i-9i-9(-1))}}} Replace {{{i^2}}} with -1.



{{{(6+6i+18i-18)/(9+9i-9i+9)}}} Multiply.



{{{(-12+24i)/(18)}}} Combine like terms.



{{{(-12)/(18)+((24)/(18))i}}} Break up the fraction.



{{{-2/3+(4/3)i}}} Reduce.



So {{{(2+6i)/(3-3i)=-2/3+(4/3)i}}} which is now in {{{a+bi}}} form where {{{a=-2/3}}} and {{{b=4/3}}}.