# What is the trigonometric form of  (4-2i) ?

Jul 13, 2017

$2 \sqrt{5}$ $c i s$ $\left(- 0.46\right)$

#### Explanation:

To find the trigonometric form, we have to know $r$, the distance of the point from the origin, and $\theta$, the angle.

We can use the following formulas:

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

$\tan \theta = \frac{b}{a}$

$r = \sqrt{{4}^{2} + {\left(- 2\right)}^{2}}$
$r = \sqrt{20}$
$r = 2 \sqrt{5}$

$\tan \theta = - \frac{2}{4}$
$\theta = {\tan}^{-} 1 \left(- \frac{2}{4}\right)$
$\theta = - 0.46$

So, the trigonometric form is $2 \sqrt{5}$ $c i s$ $\left(- 0.46\right)$ or $2 \sqrt{5} \left(\cos \left(- 0.46\right) + i \sin \left(- 0.46\right)\right)$.

Jul 13, 2017

$2 \sqrt{5} \left(\cos \left(0.46\right) - i \sin \left(0.46\right)\right)$

#### Explanation:

$\text{to convert from "color(blue)"cartesian to trig. form}$

$\text{that is " (x,y)tor(costheta+isintheta)" use}$

•color(white)(x)r=sqrt(x^2+y^2)

•color(white)(x)theta=tan^-1(y/x)color(white)(x);-pi < theta<=pi

$\text{here "x=4" and } y = - 2$

$\Rightarrow r = \sqrt{{4}^{2} + {\left(- 2\right)}^{2}} = \sqrt{20} = 2 \sqrt{5}$

4-2i is in the fourth quadrant so we must ensure that $\theta$ is in the fourth quadrant.

$\theta = {\tan}^{-} 1 \left(\frac{1}{2}\right) = 0.46 \leftarrow \textcolor{red}{\text{ related acute angle}}$

$\Rightarrow \theta = - 0.46 \leftarrow \textcolor{red}{\text{ in fourth quadrant}}$

$\Rightarrow 4 - 2 i = 2 \sqrt{5} \left(\cos \left(- 0.46\right) + i \sin \left(- 0.46\right)\right)$

$\Rightarrow 4 - 2 i = 2 \sqrt{5} \left(\cos \left(0.46\right) - i \sin \left(0.46\right)\right)$