How do you add (7+5i)+(2-4i) in trigonometric form?

Jun 13, 2018

$\sqrt{74} \cos \left(0.6202\right) + \sqrt{20} \cos \left(5.1760\right) + i \left(\sqrt{74} \sin \left(0.6202\right) + \sqrt{20} \sin \left(5.1760\right)\right)$
$9 + i$

Explanation:

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

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

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

${r}_{1} = \sqrt{{7}^{2} + {5}^{2}} = \sqrt{49 + 25} = \sqrt{74}$
${r}_{2} = \sqrt{{2}^{2} + {4}^{2}} = \sqrt{4 + 16} = \sqrt{20}$

${\theta}_{1} = {\tan}^{-} 1 \left(\frac{5}{7}\right) \approx {0.6202}^{c}$
${\theta}_{2} = {\tan}^{-} 1 \left(- 2\right) \approx - {1.1071}^{c}$

However, as $2 - 4 i$ is in quadrant 4, we need to add $2 \pi$ to get a positive angle variant.

${\theta}_{2} = 2 \pi + {\tan}^{-} 1 \left(- 2\right) \approx {5.1760}^{c}$

$\sqrt{74} \cos \left(0.6202\right) + \sqrt{20} \cos \left(5.1760\right) + i \left(\sqrt{74} \sin \left(0.6202\right) + \sqrt{20} \sin \left(5.1760\right)\right)$
$7 + 2 + i \left(5 - 4\right)$
$9 + i$

Proof:

$7 + 5 i + 2 - 4$
$\left(7 + 2\right) + i \left(5 - 4\right)$
$9 + i$