# How do you write 3i in trigonometric form?

Jun 8, 2016

I found: $z = 3 i = 3 \left[\cos \left({90}^{\circ}\right) + i \sin \left({90}^{\circ}\right)\right]$

#### Explanation:

We can find the position of your number on the complex plane:

We have that:
$z = 3 i = \rho \left[\cos \left(\theta\right) + i \sin \left(\theta\right)\right]$ in trigonometric form, where:
$\rho =$ modulus, the distance from the origin to your point; in this case: $\rho = 3$
$\theta =$ angle with the horizontal Real positive semi-axis; in this case: $\theta = {90}^{\circ}$.
So you get:
$z = 3 i = 3 \left[\cos \left({90}^{\circ}\right) + i \sin \left({90}^{\circ}\right)\right]$