Let us first write (6i-4)(6i−4) and (-3i-5)(−3i−5) in trigonometric form.
a+iba+ib can be written in trigonometric form re^(itheta)=rcostheta+irsintheta=r(costheta+isintheta)reiθ=rcosθ+irsinθ=r(cosθ+isinθ),
where r=sqrt(a^2+b^2)r=√a2+b2 and tantheta=b/atanθ=ba or theta=arctan(b/a)θ=arctan(ba)
Hence 6i-4=(-4+6i)=sqrt((-4)^2+6^2)[cosalpha+isinalpha]6i−4=(−4+6i)=√(−4)2+62[cosα+isinα] or
sqrt52e^(ialpha)√52eiα, where tanalpha=6/(-4)=-3/2tanα=6−4=−32 and
-3i-5=(-5-3i)=sqrt((-5)^2+(-3)^2)[cosbeta+isinbeta]−3i−5=(−5−3i)=√(−5)2+(−3)2[cosβ+isinβ] or
sqrt34e^(ibeta]√34eiβ, where tanbeta=(-3)/(-5)=3/5tanβ=−3−5=35
Hence (6i-4)/(-3i-5)=(sqrt52e^(ialpha))/(sqrt34e^(ibeta])=sqrt(26/17)e^(i(alpha-beta))=sqrt(28/17)(cos(alpha-beta)+isin(alpha-beta))6i−4−3i−5=√52eiα√34eiβ=√2617ei(α−β)=√2817(cos(α−β)+isin(α−β))
Now, tan(alpha-beta)=(tanalpha-tanbeta)/(1+tanalphatanbeta)tan(α−β)=tanα−tanβ1+tanαtanβ
= (-3/2-3/5)/(1+(-3/2)*(3/5))=(-21/10)/(1-9/10)=-21/10*10/1=-21−32−351+(−32)⋅(35)=−21101−910=−2110⋅101=−21
Hence (6i-4)/(-3i-5)=sqrt(26/17)(cosrho+isinrho)6i−4−3i−5=√2617(cosρ+isinρ) where rho=tan^(-1)21ρ=tan−121
