Question 372532
{{{ (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)=(8-2+17i)/(1+4)}}}
{{{ (8+i)/(1-2i)=(6+17i)/(5)}}}
{{{ (8+i)/(1-2i)=highlight((6/5)+(17/5)i)}}}