Question 167179
They probably want you to get rid of "i" in the denominator

Multiply by the conjugate of (5+i) which is (5-i) over itself, (same as mult by 1)
{{{((13i))/((5+i))}}} * {{{((5-i))/((5-i))}}}
:
Multiply the numerators and FOIL the denominators. Remember i^2 = -1
{{{((13i)*(5-i))/((25 - 5i + 5i - (i^2)))}}} = {{{((65i - 13i^2))/((25 - (-1)))}}} = {{{((65i - 13(-1)))/((25 + 1))}}} = {{{((65i + 13))/(26)}}} = {{{(65i)/26}}} + {{{13/26}}}
:
Note that we can reduce both fractions, they both are multiples of 13
{{{(5i)/2}}} + {{{1/2}}} = {{{((1 + 5i))/2}}} is the form they want probably
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