# How is -1+i√3 written in polar form?

Apr 26, 2018

See explanation.

#### Explanation:

First we need to calculate the module:

$| z | = \sqrt{{a}^{2} + {b}^{2}} = \sqrt{{\left(- 1\right)}^{2} + {\sqrt{3}}^{2}} = \sqrt{1 + 3} = 2$

Now we have:

## $z = | z | \cdot \left(\cos \varphi + i \sin \varphi\right)$

Here $| z | = 2$, so

## $\cos \varphi = \frac{R e \left(z\right)}{|} z | = - \frac{1}{2}$

The angle, for which $\cos \varphi = - \frac{1}{2}$ is $\varphi = {120}^{o}$, so now we can write the complex number in trigonometric form:

$z = 2 \cdot \left(\cos {120}^{o} + i \sin {120}^{o}\right)$

Apr 26, 2018

In polar form expressed as $2 \left(\cos 2.0944 + i \sin 2.0944\right)$

#### Explanation:

Let Z=a+i b ; Z=-1+ i sqrt 3 ; a= -1 ,b =sqrt 3  ;

$Z = - 1 + i \sqrt{3}$ is in 2nd quadrant

Modulus $| Z | = \sqrt{{a}^{2} + {b}^{2}} = \left(\sqrt{{\left(- 1\right)}^{2} + {\left(\sqrt{3}\right)}^{2}}\right) = 2$

$\tan \alpha = | \frac{b}{a} | = \left(\frac{\sqrt{3}}{-} 1\right) = \sqrt{3} \therefore \alpha = {\tan}^{-} 1 \left(\sqrt{3}\right)$

or  alpha~~ 1.047198; theta is on $2$ nd quadrant

$\therefore \theta = \pi - \alpha = 2.094395 \therefore$ Argument : $\theta \approx 2.0944$

In polar form expressed as

$| Z | \cdot \left(\cos \theta + i \sin \theta\right) \mathmr{and} 2 \left(\cos 2.0944 + i \sin 2.0944\right)$[Ans]