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#
