How do you divide #( 2i -7) / (- 5 i -8 )# in trigonometric form?

1 Answer
Apr 25, 2018

#0.51-0.58i#

Explanation:

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#