How do you convert 9i  to polar form?

Apr 11, 2018

color(blue)(9[cos(pi/2)+isin(pi/2)]

Explanation:

For complex numbers given in the form:

$a + b i$

The polar form will be:

$z = r \left[\cos \left(\theta\right) + i \sin \left(\theta\right)\right]$

Where:

$r = \sqrt{{a}^{2} + {b}^{2}}$

$\theta = \arctan \left(\frac{b}{a}\right)$

$0 + 9 i$

$r = \sqrt{{\left(0\right)}^{2} + {\left(9\right)}^{2}} = 9$

$\theta = \arctan \left(\frac{9}{0}\right)$

This is undefined, which tells us that $\theta$ is either $\frac{\pi}{2}$ or $\frac{5 \pi}{2}$.

$9 i$ is positive, so this must be in the I and II quadrants. $\theta$ is therefore $\frac{\pi}{2}$

Plugging these values into:

$z = r \left[\cos \left(\theta\right) + i \sin \left(\theta\right)\right]$

z=color(blue)(9[cos(pi/2)+isin(pi/2)]