Question 198082
{{{(3+i)/(2+3i)}}} Start with the given expression.



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



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



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



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



{{{(6-9i+2i-3i^2)/(4-6i+6i-9i^2)}}} Multiply.



{{{(9-7i)/(13)}}} Combine like terms.



{{{(9)/(13)+((-7)/(13))i}}} Break up the fraction.



{{{9/13-(7/13)i}}} Reduce.



So {{{(3+i)/(2+3i)=9/13-7/13i}}}.



So the expression is now in standard form {{{a+bi}}} where {{{a=9/13}}} and {{{b=-7/13}}}