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

Question

1837 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!

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