Question

How do I find the trigonometric form of the complex number `3i`?

Answer 1

An easy way to find any trigonometric for is by using complex numbers norms and the equation `sin²(theta) + cos²(theta) = 1`.

Choosing a generic complex `a + bi`, we find its trigonometric form by dividing `a` for the numbers norm (`sqrt(a² + b²)`) which will result in the cosine of the `theta` angle that the number refers to.

`a + bi = (sqrt(a²+b²))*cis(arccos(a/sqrt(a²+b²)))`
`:.`
The trigonometric form of `3i` is:
`0 + 3i = (sqrt(0² + 3²))*cis(arccos(0/sqrt(0²+3²))) =`
`= 3*cis(arccos(0))`
Then, `3i` in the trigonometric form is writen as `3*cis(pi/2)`.

Hope it helps.