Question

How do you divide `( 3i-1) / (8i -4 )` in trigonometric form?

Answer 1

`1/(2sqrt2)xxe^(i(alpha-beta)`, where `alpha=tan^-1(-3)` and `beta=tan^-1(-2)`

Explanation:

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

Hence, `(3i-1)=(-1+3i)=sqrt(1^2+3^2)(e^(itan^-1(-3/1)))=sqrt10e^(itan^-1(-3))`

Similarly, `(8i-4)=(-4+8i)=sqrt(4^2+8^2)(e^(itan^-1(-8/4)))=sqrt80e^(itan^-1(-2))`

Hence `(3i-1)/(8i-4)=(sqrt10e^(itan^-1(-3)))/(sqrt80e^(itan^-1(-2)))`

or `sqrt(1/8)xxe^(i(tan^-1(-3))-(tan^-1(-2))` or

`1/(2sqrt2)xxe^(i(alpha-beta)`, where `alpha=tan^-1(-3)` and `beta=tan^-1(-2)`