Question 159832
{{{ (3+4i)/(6-2i) }}} Start with the given expression


{{{ ((3+4i)/(6-2i))((6+2i)/(6+2i)) }}} Multiply both numerator and denominator by conjugate of the denominator {{{6-2i}}}



{{{ ((3+4i)(6+2i))/((6-2i)(6+2i))}}} Combine the fractions



{{{ (18+30i+8i^2)/(36-4i^2)}}} FOIL



{{{ (18+30i-8)/(36+4)}}} Replace {{{i^2}}} with -1 and multiply



{{{ (10+30i)/(40)}}} Combine like terms.



{{{10/40+(30/40)i}}} Break up the fraction



{{{1/4+(3/4)i}}} Reduce



So {{{ (3+4i)/(6-2i) }}}  simplifies to {{{1/4+(3/4)i}}}. In other words, {{{ (3+4i)/(6-2i) = 1/4+(3/4)i}}}. Notice how the result is in standard form {{{a+bi}}} where {{{a=1/4}}} and {{{b=3/4}}}