# How do you add (5+5i)+(-3+i) in trigonometric form?

Nov 8, 2017

$\left(5 + 5 i\right) + \left(- 3 + i\right) = 2 \sqrt{10} \left(\cos \left(\frac{5}{4}\right) + i \sin \left(\frac{5}{4}\right)\right)$

#### Explanation:

$\left(5 + 5 i\right) + \left(- 3 + i\right) = \left(5 - 3\right) + \left(5 i + i\right) = 2 + 6 i$

Any complex number of the form $a + b i , a , b \in \mathbb{R}$ may be written in its mod-arg form $r \left(\cos \vartheta + i \sin \vartheta\right)$ where $r = \sqrt{{a}^{2} + {b}^{2}}$ and $\vartheta = \arctan \left(b \text{/} a\right)$

$r = \sqrt{{2}^{2} + {6}^{2}} = \sqrt{40} = 2 \sqrt{10}$

$\vartheta = \arctan \left(\frac{6}{2}\right) = \arctan 3 \approx \frac{5}{4}$

$\therefore 2 + 6 i = 2 \sqrt{10} \left(\cos \left(\frac{5}{4}\right) + i \sin \left(\frac{5}{4}\right)\right)$