In trigonometric form,
i-3=sqrt(10)(cosalpha+isinalpha)i−3=√10(cosα+isinα), where
cosalpha=-3/sqrt(10), sinalpha=1/sqrt(10)cosα=−3√10,sinα=1√10
2i+1=sqrt(5)(cosbeta+isinbeta)2i+1=√5(cosβ+isinβ)
(cosbeta=1/sqrt(5), sinbeta=2/sqrt(5)cosβ=1√5,sinβ=2√5).
So,
(i-3)/(2i+1)=(sqrt(10)(cosalpha+isinalpha))/(sqrt(5)(cosbeta+isinbeta)i−32i+1=√10(cosα+isinα)√5(cosβ+isinβ)
=sqrt(2)(cos(alpha-beta)+isin(alpha-beta))=√2(cos(α−β)+isin(α−β)).
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.
