Question 207791
{{{(2)/(3-i)}}} Start with the given expression.



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



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



{{{((2)(3)+(2)(i))/((3-i)(3+i))}}} Distribute



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



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



{{{(6+2i)/(9+3i-3i-(-1))}}} Replace {{{i^2}}} with -1



{{{(6+2i)/(9+3i-3i+1)}}} Simplify



{{{(6+2i)/(10)}}} Combine like terms.



{{{(6)/(10)+((2)/(10))i}}} Break up the fraction.



{{{3/5+(1/5)i}}} Reduce.



So {{{(2)/(3-i)=3/5+(1/5)i}}}.



So the expression is now in standard form {{{a+bi}}} where {{{a=3/5}}} and {{{b=1/5}}}