# How do you add (-5-2i)+(7-6i) in trigonometric form?

Jul 15, 2017

$\left(- 5 - 2 i\right) + \left(7 - 6 i\right) = 2 \sqrt{17} \left(\cos 1.33 - i \sin 1.33\right)$

#### Explanation:

$\left(- 5 - 2 i\right) + \left(7 - 6 i\right) = 2 - 8 i$

This can be written in trigonometric form as:

$a + b i = r \left(\cos \vartheta + i \sin \vartheta\right)$ where $r = \sqrt{{a}^{2} + {b}^{2}}$ and $\vartheta = \arctan \left(\frac{y}{x}\right)$

$\sqrt{{2}^{2} + {\left(- 8\right)}^{2}} = 2 \sqrt{17}$

$\arctan \left(- \frac{8}{2}\right) = - 1.33$

$\therefore 2 - 8 i = 2 \sqrt{17} \left(\cos \left(- 1.33\right) + i \sin \left(- 1.33\right)\right)$

Using the properties of the $\sin$ and $\cos$ functions, this can be rewritten as:

$2 - 8 i = 2 \sqrt{17} \left(\cos 1.33 - i \sin 1.33\right)$