(-6i+5)/(i+2)=f(r,theta)−6i+5i+2=f(r,θ)
z_1 = -6i+5z1=−6i+5
r_1 = sqrt((-6)^2+5^2) =sqrt(36+25)=sqrt61r1=√(−6)2+52=√36+25=√61
theta_1=tan^-1 (5/-6)=2pi-tan^-1(5/6)θ1=tan−1(5−6)=2π−tan−1(56)
z_2=i+2z2=i+2
r_2 = sqrt(1^2+2^2) =sqrt(1+4)=sqrt5r2=√12+22=√1+4=√5
theta_2=tan^-1 (2/1)=2pi+tan^-1 2θ2=tan−1(21)=2π+tan−12
Thus,
(-6i+5)/(i+2)=(sqrt61, (2pi-tan^-1(5/6)))/(sqrt5, (2pi+tan^-1 2))−6i+5i+2=√61,(2π−tan−1(56))√5,(2π+tan−12)
By De-Moivre's Theorem,
(sqrt61, (2pi-tan^-1(5/6)))/(sqrt5, (2pi+tan^-1 2))√61,(2π−tan−1(56))√5,(2π+tan−12)
=sqrt 61 /(sqrt 5) xxcis(2pi-tan^-1(5/6)-(2pi+tan^-1 2))=√61√5×cis(2π−tan−1(56)−(2π+tan−12))
=sqrt(61/5)cis(-tan^-1(5/6)-tan^-1 2)=√615cis(−tan−1(56)−tan−12)
=sqrt(61/5)cis(-(tan^-1(5/6)+tan^-1 2))=√615cis(−(tan−1(56)+tan−12))
tan^-1 (5/6) +tan^-1 2 = tan^-1 ((5/6+2)/(1-5/6xx2))tan−1(56)+tan−12=tan−1(56+21−56×2)
=tan^-1((5xx1+2xx6)/(1xx6-5xx3))=tan^-1 ((5+12)/(6-15))=tan−1(5×1+2×61×6−5×3)=tan−1(5+126−15)
=tan^-1(17/-9)=tan−1(17−9)
Thus,
(-6i+5)/(i+2)=sqrt(61/5)cis(-tan^-1(17/-9))−6i+5i+2=√615cis(−tan−1(17−9))
