# How do you add (1-5i)+(-2+i) in trigonometric form?

$\sqrt{17} \left[\cos \left({\tan}^{-} 1 \left(\frac{- 4}{- 1}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{- 4}{- 1}\right)\right)\right]$
$\sqrt{17} \left[\cos \left(4.4674103172578\right) + i \sin \left(4.4674103172578\right)\right] \text{ }$Radian

$\sqrt{17} \left[\cos \left({255.96375653207}^{\circ}\right) + i \sin \left({255.96375653207}^{\circ}\right)\right]$

#### Explanation:

First we add the complex numbers
$\left(1 - 5 i\right) + \left(- 2 + i\right) =$

$= \left(1 - 2\right) + \left(- 5 i + i\right)$

$= - 1 - 4 i$

Convert to trigonometric form

for complex number $a + i b$

$a + i b = \sqrt{{a}^{2} + {b}^{2}} \cdot \left[\cos \left({\tan}^{-} 1 \left(\frac{b}{a}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{b}{a}\right)\right)\right]$

so we let $a = - 1$ and $b = - 4$

$- 1 - 4 i =$
$\sqrt{{\left(- 1\right)}^{2} + {\left(- 4\right)}^{2}} \cdot \left[\cos \left({\tan}^{-} 1 \left(\frac{- 4}{- 1}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{- 4}{- 1}\right)\right)\right]$

In the complex rectangular coordinate system, this is located at the 3rd quadrant

$\sqrt{17} \left[\cos \left({\tan}^{-} 1 \left(\frac{- 4}{- 1}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{- 4}{- 1}\right)\right)\right]$
$\sqrt{17} \left[\cos \left(4.4674103172578\right) + i \sin \left(4.4674103172578\right)\right] \text{ }$Radian

$\sqrt{17} \left[\cos \left({255.96375653207}^{\circ}\right) + i \sin \left({255.96375653207}^{\circ}\right)\right]$

have a nice day !