Question 683232
{{{(8+i)/(1-2i)}}}
multiply by the conjugate of the denominator over itself
{{{(8+i)/(1-2i)}}} * {{{(1+2i)/(1+2i)}}} = {{{((8+i)(1+2i))/(1 - 4i^2)}}}
FOIL
{{{((8 + 16i + 1 +2i^2))/(1 - 4(-1))}}} = {{{((8+17i+2(-1)))/(1 + 4)}}} = {{{(8+17i-2)/5}}} = {{{(6+17i)/5}}}
or
{{{6/5}}} + {{{17/5}}}i