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

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 $a + b i$, 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)))
$\therefore$
The trigonometric form of $3 i$ is:
0 + 3i = (sqrt(0² + 3²))*cis(arccos(0/sqrt(0²+3²))) =
$= 3 \cdot c i s \left(\arccos \left(0\right)\right)$
Then, $3 i$ in the trigonometric form is writen as $3 \cdot c i s \left(\frac{\pi}{2}\right)$.

Hope it helps.