# How do you express the complex number in trigonometric form: -7-5i?

Aug 23, 2017

$- 7 - 5 i = \sqrt{74} \left(\cos 2.52 - i \sin 2.52\right)$

#### Explanation:

Any $a + b i$ can be rewritten in mod-arg form as $r \left(\cos \vartheta + i \sin \vartheta\right)$ where $r = \sqrt{{a}^{2} + {b}^{2}}$ and $\vartheta = \arctan \left(\frac{b}{a}\right)$

$| - 7 - 5 i | = \sqrt{{5}^{2} + {7}^{2}} = \sqrt{74}$

$\text{arg} \left(- 7 - 5 i\right) = \arctan \left(\frac{- 5}{- 7}\right) = - 2.52$

$\therefore - 5 - 7 i \equiv \sqrt{74} \left(\cos \left(- 2.52\right) + i \sin \left(- 2.52\right)\right) = \sqrt{74} \left(\cos 2.52 - i \sin 2.52\right)$

Aug 23, 2017

$\sqrt{74} \left(\cos \left(2.52\right) - i \sin \left(2.52\right)\right)$

#### Explanation:

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

$\text{that is "x+yitor(costheta+isintheta)" using}$

•color(white)(x)r=sqrt(x^2+y^2)

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

$\text{here "x=-7" and } y = - 5$

$\Rightarrow r = \sqrt{{\left(- 7\right)}^{2} + {\left(- 5\right)}^{2}} = \sqrt{74}$

$- 7 - 5 i \text{ is in the third quadrant so we must ensure } \theta$
$\text{is in the third quadrant}$

$\theta = {\tan}^{-} 1 \left(\frac{5}{7}\right) = 0.62 \leftarrow \textcolor{b l u e}{\text{ related acute angle}}$

$\Rightarrow \theta = - \pi + 0.62 = - 2.52 \leftarrow \textcolor{b l u e}{\text{ in third quadrant}}$

$\Rightarrow - 7 - 5 i \to \sqrt{74} \left(\cos \left(- 2.52\right) + i \sin \left(- 2.52\right)\right)$

$= \sqrt{74} \left(\cos \left(2.52\right) - i \sin \left(2.52\right)\right)$