How do you divide #( 9i- 7) / (- 5 i -6 )# in trigonometric form?

1 Answer
1s2s2p
Nov 8, 2017

#sqrt(130)/sqrt(61)cos(-91.931)+isin(-91.931)#

Explanation:

#z=a+bi# in trigonometric form is #z=r(cos\theta+isin\theta)#, where #r=sqrt(a^2+b^2)# and #\theta=tan^(-1)(b/a)#

So, we have #(-7+9i)/(-6-5i)=(sqrt((-7)^2+9^2)(cos(tan^(-1)(9/-7))+isin(tan^(-1)(9/-7))))/(sqrt((-6)^2+(-5)^2)(cos(tan^(-1)((-5)/-6))+isin(tan^(-1)((-5)/-6))))#
#=(sqrt(130)(cos(-52.125)+isin(-52.125)))/(sqrt(61)(cos(39.806)+isin(39.806)))#
#=sqrt(130)/sqrt(61)cos(-52.125-39.806)+isin(-52.125-39.806)#
#=sqrt(130)/sqrt(61)cos(-91.931)+isin(-91.931)#

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