# How do you add (-8+4i) and (-3-3i) in trigonometric form?

$\sqrt{122} \left[\cos \left({\tan}^{-} 1 \left(\frac{1}{- 11}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{1}{- 11}\right)\right)\right] \text{ " }$OR
$\sqrt{122} \left[\cos \left({174.806}^{\circ}\right) + i \sin \left({174.806}^{\circ}\right)\right] \text{ " }$

#### Explanation:

Add $\left(- 8 + 4 i\right)$ and $\left(- 3 - 3 i\right)$ to obtain

$- 11 + i$

Use the formula

a+bi=sqrt(a^2+b^2)[cos(tan^-1 (b/a))+i*sin(tan^-1 (b/a)]

Let $a = - 11$ and $b = 1$

magnitude$\text{ " " } r = \sqrt{{a}^{2} + {b}^{2}} = \sqrt{{\left(- 11\right)}^{2} + {1}^{2}} = \sqrt{122}$

the angle $\theta = {\tan}^{-} 1 \left(\frac{1}{- 11}\right) = {174.806}^{\circ}$

$\sqrt{122} \left[\cos \left({\tan}^{-} 1 \left(\frac{1}{- 11}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{1}{- 11}\right)\right)\right] \text{ " }$OR
$\sqrt{122} \left[\cos \left({174.806}^{\circ}\right) + i \sin \left({174.806}^{\circ}\right)\right] \text{ " }$

have a nice day... from the Philippines..