We have z=(-7+2i)/(-8-5i)=(7-2i)/(8+5i)
For z=a+bi, z=r(costheta+isintheta), where:
- r=sqrt(a^2+b^2)
- theta=tan^-1(b/a)
For 7-2i:
r=sqrt(7^2+2^2)=sqrt53
theta=tan^-1(-2/7)~~-0.28^c, however 7-2i is in quadrant 4 and so must add 2pi to it to make it positive, also 2pi would be going around a circle back.
theta=tan^-1(-2/7)+2pi~~6^c
For 8+5i:
r=sqrt(8^2+5^2)=sqrt89
theta=tan^-1(5/8)~~0.56^c
When we have z_1/z_1 in trig form, we do r_1/r_1(cos(theta_1-theta_2)+isin(theta_1-theta_2)
z_1/z_2=sqrt53/sqrt89(cos(6-0.56)+isin(6-0.56))=sqrt4717/89(cos(
5.44)+isin(5.44))=0.51-0.58i
Proof:
(7-2i)/(8+5i)*(8-5i)/(8-5i)=(56-51i-10)/(64+25)=(46-51i)/89=0.52-0.57
