Let us first write (2i-6)(2i−6) and (-12i+4)(−12i+4) 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 2i-6=(-6+2i)=sqrt((-6)^2+2^2)[cosalpha+isinalpha]2i−6=(−6+2i)=√(−6)2+22[cosα+isinα] or
sqrt40e^(ialpha)√40eiα, where tanalpha=(-1)/3tanα=−13 and
-12i+4=(4-12i)=sqrt(4^2+(-12)^2)[cosbeta+isinbeta]−12i+4=(4−12i)=√42+(−12)2[cosβ+isinβ] or
sqrt160e^(ibeta]√160eiβ, where tanbeta=(-12)/4=-3tanβ=−124=−3
Hence (2i-6)-:(-12i+4)=(sqrt40e^(ialpha))/(sqrt160e^(ibeta])=sqrt(1/4)e^(i(alpha-beta))=e^(i(alpha-beta))/2=1/2(cos(alpha-beta)+isin(alpha-beta))(2i−6)÷(−12i+4)=√40eiα√160eiβ=√14ei(α−β)=ei(α−β)2=12(cos(α−β)+isin(α−β))
Now, tan(alpha-beta)=(tanalpha-tanbeta)/(1+tanalphatanbeta)tan(α−β)=tanα−tanβ1+tanαtanβ
= (-1/3+(-3))/(1+(-1/3)*(-3))=(-10/3)/2=-5/3−13+(−3)1+(−13)⋅(−3)=−1032=−53
Hence (2i-6)-:(-12i+4)=1/2(cosrho+isinrho)(2i−6)÷(−12i+4)=12(cosρ+isinρ) where rho=tan^(-1)(-5/3)ρ=tan−1(−53)
