Question 866713
{{{(8+i)/(1-2i)=((8+i)/(1-2i))*((1+2i)/(1+2i))}}}
{{{(8+i)/(1-2i)=(8+16i+i+2i^2)/(1+2i-2i-4i^2)}}}
{{{(8+i)/(1-2i)=(8+17i-2)/(1+4)}}}
{{{(8+i)/(1-2i)=(6+17i)/(5)}}}