Question

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

Answer 1

`z=a+bi` = `r[z]e^(i*arctan(b/a))` by definition,
for `a,b,r[z] in RR`, `i = sqrt(-1)` , and `r[z] = sqrt(a^2+b^2)`
so for complex numbers `u=v+iw and z=a+bi`,
`(u/z)` =`(r[u])/(r[z]]` `e^(i(arctan(w/v)-arctan(b/a))`

In this case:
`u=2i-9` = `sqrt(2^2+9^2)*e^(i*arctan(-2/9))=9.2195e^(-.2187i)`
`z=7-7i = sqrt(2*(7^2))*e^(i*arctan(-1))=9.899e^(-.25ipi)`

So:
`((2i-9)/(-7i+7))` = `(9.2195/9.899)` `e^(i*(-.2187-.25*pi)`
= `0.9314*e^(-1.0041i)`