# How do you add (7+4i) and (2+6i) in trigonometric form?

Jan 29, 2018

$9 + 10 i$

#### Explanation:

If we have a complex number such that $z = a + b i$, then $z = r \left(\cos \theta + i \sin \theta\right)$, where:

• $r = \sqrt{{a}^{2} + {b}^{2}}$
• $\theta = {\tan}^{- 1} \left(\frac{b}{a}\right)$

For ${z}_{1} = 7 + 4 i$ :
$r = \sqrt{{7}^{2} + {4}^{2}} = \sqrt{49 + 16} = \sqrt{65}$
$\theta = {\tan}^{- 1} \left(\frac{4}{7}\right) \approx 29.7$
$= \sqrt{65} \left(\cos \left(29.7\right) + i \sin \left(29.7\right)\right)$

For ${z}_{2} = 2 + 6 i$ :
$r = \sqrt{{2}^{2} + {6}^{2}} = \sqrt{4 + 36} = \sqrt{40}$
$\theta = {\tan}^{- 1} \left(\frac{6}{2}\right) \approx 71.6$
$= \sqrt{40} \left(\cos \left(71.6\right) + i \sin \left(71.6\right)\right)$

For ${z}_{1} + {z}_{2}$ :
${z}_{1} + {z}_{2} = \sqrt{65} \cos \left(29.7\right) + i \sqrt{65} \sin \left(29.7\right) + \sqrt{40} \cos \left(71.6\right) + i \sqrt{40} \sin \left(71.6\right)$

$\textcolor{w h i t e}{{z}_{1} + {z}_{2}} = \sqrt{65} \cos \left(29.7\right) + \sqrt{40} \cos \left(71.6\right) + i \left(\sqrt{65} \sin \left(29.7\right) + \sqrt{40} \sin \left(71.6\right)\right)$

$\textcolor{w h i t e}{{z}_{1} + {z}_{2}} = 8.999470954 + 9.995734316 i$#

$\textcolor{w h i t e}{{z}_{1} + {z}_{2}} \approx 9 + 10 i$

Proof:
$7 + 4 i + 2 + 6 i = \left(7 + 2\right) + i \left(4 + 6\right) = 9 + 10 i$