How do you add (1+7i)+(-4-2i) in trigonometric form?

Jun 25, 2018

color(brown)(=> -3 + 5i

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(1^2+7^2))=sqrt50
${r}_{2} = \sqrt{- {4}^{2} \pm {2}^{2}} = \sqrt{20}$

${\theta}_{1} = {\tan}^{-} 1 \left(\frac{7}{1}\right) \approx {81.87}^{\circ} , \text{ I quadrant}$
${\theta}_{2} = {\tan}^{-} 1 \left(- \frac{2}{-} 4\right) \approx {206.57}^{\circ} , \text{ III quadrant}$

${z}_{1} + {z}_{2} = \sqrt{50} \cos \left(81.87\right) + \sqrt{20} \cos \left(206.57\right) + i \left(\sqrt{50} \sin \left(81.87\right) + \sqrt{20} \sin \left(206.57\right)\right)$

$\implies 1 - 4 + i \left(7 - 2\right)$
color(brown)(=> -3 + 5i

Proof:

$1 + 7 i - 4 - 2 i$

$\implies - 3 + 5 i$