Question 189902
{{{(1)/(3+i)}}} Start with the given expression.



{{{((1)/(3+i))((3-i)/(3-i))}}} Multiply the fraction by {{{(3-i)/(3-i)}}}.



{{{((1)(3-i))/((3+i)(3-i))}}} Combine the fractions.



{{{(3-i)/((3+i)(3-i))}}} Distribute.



{{{(3-i)/(9-3i+3i-i^2)}}} FOIL the denominator.



{{{(3-i)/(9-3i+3i-(-1))}}} Replace {{{i^2}}} with -1



{{{(3-i)/(9-3i+3i+1)}}} Simplify



{{{(3-i)/(10)}}} Combine like terms.



{{{(3)/(10)+((-1)/(10))i}}} Break up the fraction.



{{{3/10-(1/10)i}}} Reduce.



So {{{(1)/(3+i)=3/10-(1/10)i}}}.



So the expression is now in standard form {{{a+bi}}} where {{{a=3/10}}} and {{{b=-1/10}}}