# How do you add (-2+3i)+(6+i) in trigonometric form?

Jun 25, 2018

color(crimson)((-2 + 3 i) + (6 + i) = 4 + 4 i = 4(1 + i)

#### Explanation:

$z = a + b i = r \left(\cos \theta + i \sin \theta\right)$

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

${r}_{1} \left(\cos \left({\theta}_{1}\right) + i \sin \left({\theta}_{2}\right)\right) + {r}_{2} \left(\cos \left({\theta}_{2}\right) + i \sin \left({\theta}_{2}\right)\right) = {r}_{1} \cos \left({\theta}_{1}\right) + {r}_{2} \cos \left({\theta}_{2}\right) + i \left({r}_{1} \sin \left({\theta}_{1}\right) + {r}_{2} \sin \left({\theta}_{2}\right)\right)$

r_1=sqrt(-2^2+3^2))=sqrt 13
${r}_{2} = \sqrt{{6}^{2} \pm {1}^{2}} = \sqrt{37}$

${\theta}_{1} = {\tan}^{-} 1 \left(\frac{3}{-} 2\right) \approx {123.69}^{\circ} , \text{ II quadrant}$
${\theta}_{2} = {\tan}^{-} 1 \left(\frac{1}{6}\right) \approx {9.4623}^{\circ} , \text{ I quadrant}$

${z}_{1} + {z}_{2} = \sqrt{13} \cos \left(123.69\right) + \sqrt{37} \cos \left(9.4623\right) + i \left(\sqrt{13} \sin \left(123.69\right) + \sqrt{37} \sin \left(9.4623\right)\right)$

$\implies - 2 + 6 + i \left(3 + 1\right)$
color(brown)(=> 4 + 4 i

Proof:

$- 2 + 3 i + 6 + i$

$\implies 4 + 4 i$