First I will rewrite the expressions in the form of a+bi
(3+i)/(7-3i)
For a complex number z=a+bi, z=r(costheta+isintheta), where:
- r=sqrt(a^2+b^2)
- theta=tan^-1(b/a)
Let's call 3+i z_1 and 7-3i z_2.
For z_1:
z_1=r_1(costheta_1+isintheta_1)
r_1=sqrt(3^2+1^2)=sqrt(9+1)=sqrt(10)
theta_1=tan^-1(1/3)=0.32^c
z_1=sqrt(10)(cos(0.32)+isin(0.32))
For z_2:
z_2=r_2(costheta_2+isintheta_2)
r_2=sqrt(7^2+(-3)^2)=sqrt(58)
theta_2=tan^-1(-3/7)=-0.40^c
However, since 7-3i is in quadrant 4, we need to get a positive angle equivalent (the negative angle goes clockwise around the circle, and we need an anticlockwise angle).
To get a positive angle equivalent, we add 2pi, tan^-1(-3/7)+2pi=5.88^c
z_2=sqrt(58)(cos(5.88)+isin(5.88))
For z_1/z_2:
z_1/z_2=r_1/r_2(cos(theta_1-theta_2)+isin(theta_1-theta_2))
color(white)(z_1/z_2)=sqrt(10)/sqrt(58)(cos[tan^-1(1/3)-(tan^-1(-3/7)+2pi)]+isin[tan^-1(1/3)-(tan^-1(-3/7)+2pi)])
color(white)(z_1/z_2)=sqrt(145)/29(cos[tan^-1(1/3)-tan^-1(-3/7)-2pi]+isin[tan^-1(1/3)-tan^-1(-3/7)-2pi])
color(white)(z_1/z_2)=sqrt(145)/29(cos(-5.56)+isin(-5.56))
color(white)(z_1/z_2)=sqrt(145)/29cos(-5.56)+isqrt(145)/29sin(-5.56)
color(white)(z_1/z_2)=0.311+0.275i
Proof:
(3+i)/(7-3i)*(7+3i)/(7+3i)=((3+i)(7+3i))/((7-3i)(7+3i))=(21+7i+9i+3i^2)/(49+21i-21i-9i^2)=(21+16i+3i^2)/(49-9i^2)
i^2=-1
=(21+16i-3)/(49+9)=(18+16i)/58=9/29+8/29i~~0.310+0.275i
