How do you divide # (7-9i)/(-2-9i) # in trigonometric form?

Question

1312 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-21T04:50:34

#sqrt(442)/17[cos(tan^-1((-81)/-67))+i*sin(tan^-1((-81)/-67))]# OR

#sqrt(442)/17[cos(50.403791360249^@)+i*sin(50.403791360249^@)]#

Convert to Trigonometric forms first

#7-9i=sqrt130[cos(tan^-1((-9)/7))+i sin(tan^-1((-9)/7))]#

#-2-9i=sqrt85[cos(tan^-1((-9)/-2))+i sin(tan^-1((-9)/-2))]#

Divide equals by equals

#(7-9i)/(-2-9i)=#

#(sqrt130/sqrt85)[cos(tan^-1((-9)/7)-tan^-1((-9)/-2))+i sin(tan^-1((-9)/7)-tan^-1((-9)/-2))]#

Take note of the formula:

#tan (A-B)=(Tan A-Tan B)/(1+Tan A* Tan B)#

also

#A-B=Tan^-1 ((Tan A-Tan B)/(1+Tan A* Tan B))#

#sqrt(442)/17[cos(tan^-1((-81)/-67))+i*sin(tan^-1((-81)/-67))]#

have a nice day!