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

Dec 21, 2014

$z = 3 \left[\cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right)\right] = 3 i \sin \left(\frac{\pi}{2}\right)$

When you have to convert a complex number, given in "rectangular form" ( $z = a + i b$ ), to trigonometric form $z = r \left[\cos \left(\theta\right) + i \sin \left(\theta\right)\right]$ you need to evaluate:
1) the modulus $r$ (using Pitagora's Theorem);
2) the argument $\theta$ (using trigonometry).

Graphically:

In your case you have: $z = 0 + 3 i = 3 i$ so that:
1) $r = \sqrt{{3}^{2} + {0}^{2}} = 3$
2) $\theta = \arctan \left(\frac{3}{0}\right) = \frac{\pi}{2}$

Graphically: