Question

How do you divide `(6+9i)/(1+i)` in trigonometric form?

Answer 1

Sorry for the delay. Below is my answer.

Explanation:

First step, get rid of complex denominator by multiplying top and bottom by its complex conjugate
`r = (6 + 9i)(1 - i)/[(1 + i)(1-i)] = (6 + 9i - 6i + 9)/2 = 15/2 + 3i/2`

Next you factor out the modulus or norm of the complex number. and write

`r = sqrt(225 + 9)/2(cos alpha + i sin alpha)`

where `alpha` is such that `tan alpha = 1/5`