# How do you write the complex number -2i in polar form?

##### 2 Answers
Mar 23, 2017

Please see the explanation.

#### Explanation:

Because the real part (a) of the complex number is zero, you cannot use $\theta = {\tan}^{-} 1 \left(\frac{b}{a}\right)$; you must know that the angle is either $\frac{\pi}{2}$ or $3 \frac{\pi}{2}$. Because the sign of the complex part is negative, you must know that this makes the angle the latter, $3 \frac{\pi}{2}$.

$\theta = 3 \frac{\pi}{2}$

You can see that the magnitude is 2.

The polar form is:

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

Mar 23, 2017

I hope a get it right: $2 \cdot \left(\cos \left(270\right) + i \sin \left(270\right)\right)$

#### Explanation:

$r = \sqrt{{x}^{2} + {y}^{2}} = \sqrt{4} = 2$
$\alpha = c t {g}^{- 1} \left(\frac{0}{-} 2\right) = 90$
$\alpha = 360 - 90 = 270$

In polar:
$2 \cdot \left(\cos \left(270\right) + i \sin \left(270\right)\right)$