In trigonometric form,
#i-3=sqrt(10)(cosalpha+isinalpha)#, where
#cosalpha=-3/sqrt(10), sinalpha=1/sqrt(10)#
#2i+1=sqrt(5)(cosbeta+isinbeta)#
(#cosbeta=1/sqrt(5), sinbeta=2/sqrt(5)#).
So,
#(i-3)/(2i+1)=(sqrt(10)(cosalpha+isinalpha))/(sqrt(5)(cosbeta+isinbeta)#
#=sqrt(2)(cos(alpha-beta)+isin(alpha-beta))#.
Then, use the trigonometric addition formulas.
#cos(α-β)=cosαcosbeta+sinalphasinbeta#
#=-3/sqrt(10)*1/sqrt(5)+1/sqrt(10)*2/sqrt(5)#
#=-sqrt(2)/10#
#sin(alpha-beta)=sinalphacosbeta-cosalphasinbeta#
#=1/sqrt(10)*1/sqrt(5)-(-3/sqrt(10))*2/sqrt(5)#
#=(7sqrt(2))/10#
Therefore,
#(i-3)/(2i+1)=sqrt(2){-sqrt(2)/10+i*((7sqrt(2))/10)}#
#=-1/5+7/5 i#.
