jmc

Since the coefficients of the polynomial are all real, another is the conjugate of $1-2i,$ namely $1+2i.$ Let $r$ be the third root. Then the polynomial is $$(x-1+2i)(x-1-2i)(x-r)=x^3-(r+2)x^2+(2r+5)x-5r.$$Then $2r+5=-1,$ so $r=-3.$ Then $a=-(r+2)=1$ and $b=-5r=15,$ so $(a,b)=\boxed{(1,15)}.$