Question

How do you divide `(7-4i)/(3+2i)` in trigonometric form?

Answer 1

`sqrt5xxe^(i(alpha-beta)`, where `alpha=tan^-1(-4/7)` and `beta=tan^-1(2/3)`

Explanation:

`(a+bi)` an be written in as `sqrt(a^2+b^2)e^(i(tan^-1(b/a))`

Hence, `(7-4i)=sqrt(7^2+4^2)(e^(itan^-1(-4/7)))=sqrt65e^(itan^-1(-4/7))`

Similarly, `(3+2i)=sqrt(3^2+2^2)(e^(itan^-1(2/3)))=sqrt13e^(itan^-1(2/3))`

Hence `(7-4i)/(3+2i)=(sqrt65e^(itan^-1(-4/7)))/(sqrt13e^(itan^-1(2/3)))`

or `sqrt5xxe^(i(tan^-1(-4/7))-(tan^-1(2/3)))` or

`sqrt5xxe^(i(alpha-beta)`, where `alpha=tan^-1(-4/7)` and `beta=tan^-1(2/3)`