# How do you add (-8-9i)+(3-6i) in trigonometric form?

Jan 25, 2018

${z}_{1} + {z}_{2} = \left(\sqrt{145} \cos \left(\arctan \left(\frac{9}{8}\right) + \pi\right) + 3 \sqrt{5} \cos \left(\arctan \left(- 2\right) + 2 \pi\right)\right) + i \left(\sqrt{145} \sin \left(\arctan \left(\frac{9}{8}\right) + \pi\right) + 3 \sqrt{5} \sin \left(\arctan \left(- 2\right) + 2 \pi\right)\right)$

$\textcolor{w h i t e}{{z}_{1} + {z}_{2}} \approx \left(\sqrt{145} \cos 3.9857 + 3 \sqrt{5} \cos 5.1760\right) + i \left(\sqrt{145} \sin 3.9857 + 3 \sqrt{5} \sin 5.1760\right)$

$\textcolor{w h i t e}{{z}_{1} + {z}_{2}} = - 5 - 15 i$

#### Explanation:

For a complex number $a + b i$ we can repesent this in trigonometric form, as $z = r \left(\cos \theta + i \sin \theta\right)$

$r = \sqrt{{a}^{2} + {b}^{2}}$

$\theta = \arctan \left(\frac{b}{a}\right)$

So, for ${z}_{1} = - 8 - 9 i$, we can find that ${r}_{1} = \sqrt{{\left(- 8\right)}^{2} + {\left(- 9\right)}^{2}} = \sqrt{145}$

${\theta}_{1} = \arctan \left(\frac{- 9}{- 8}\right)$

However, $- 8 - 9 i$ is in quadrant 3, but $\theta$ is in quadrant 1, so we need to add $\pi$

${\theta}_{1} = \arctan \left(\frac{- 9}{- 8}\right) + \pi \approx 3.9857$

${z}_{1} \approx \sqrt{145} \left(\cos \left(3.9857\right) + i \sin \left(3.9857\right)\right)$

For ${z}_{2} = 3 - 6 i$, ${r}_{2} = \sqrt{{3}^{2} + {\left(- 6\right)}^{2}} = \sqrt{45} = 3 \sqrt{5}$

${\theta}_{2} = \arctan \left(- \frac{3}{6}\right) = \arctan \left(- 2\right)$

$3 - 6 i$ is in quadrant 4, while $\theta$ is in quadrant 4, however, we need theta to be positive, while being in the range $\theta \in \left[0 , 2 \pi\right)$. So, we must add $2 \pi$, even though this is going all the way around the circle (in terms of the 4 quadrants), we will effectively arrive at the same place.

${\theta}_{2} = \arctan \left(- \frac{3}{6}\right) = \arctan \left(- 2\right) + 2 \pi \approx 5.1760$

${z}_{2} \approx 3 \sqrt{5} \left(\cos \left(5.1760\right) + i \sin \left(5.1760\right)\right)$

We now have ${z}_{1}$ and ${z}_{2}$ in trig form.

${z}_{1} + {z}_{2} = {r}_{1} \cos {\theta}_{1} + i {r}_{1} \sin {\theta}_{1} + {r}_{2} \cos {\theta}_{2} + i {r}_{2} \sin {\theta}_{2}$

$\textcolor{w h i t e}{{z}_{1} + {z}_{2}} = \left({r}_{1} \cos {\theta}_{1} + {r}_{2} \cos {\theta}_{2}\right) + i \left({r}_{1} \sin {\theta}_{1} + {r}_{2} \sin {\theta}_{2}\right)$

$\textcolor{w h i t e}{{z}_{1} + {z}_{2}} \approx \left(\sqrt{145} \cos 3.9857 + 3 \sqrt{5} \cos 5.1760\right) + i \left(\sqrt{145} \sin 3.9857 + 3 \sqrt{5} \sin 5.1760\right)$

$\textcolor{w h i t e}{{z}_{1} + {z}_{2}} \approx - 5.000639287 + i \left(- 14.99973664\right)$

$\textcolor{w h i t e}{{z}_{1} + {z}_{2}} \approx - 5.000639287 - 14.99973664 i$

$\textcolor{w h i t e}{{z}_{1} + {z}_{2}} \approx - 5 - 15 i$

If we use ${\theta}_{1} = \arctan \left(\frac{9}{8}\right) + \pi$ and ${\theta}_{2} = \arctan \left(- 2\right) + 2 \pi$ we get:

${z}_{1} + {z}_{2} = \left(\sqrt{145} \cos \left(\arctan \left(\frac{9}{8}\right) + \pi\right) + 3 \sqrt{5} \cos \left(\arctan \left(- 2\right) + 2 \pi\right)\right) + i \left(\sqrt{145} \sin \left(\arctan \left(\frac{9}{8}\right) + \pi\right) + 3 \sqrt{5} \sin \left(\arctan \left(- 2\right) + 2 \pi\right)\right)$

$\textcolor{w h i t e}{{z}_{1} + {z}_{2}} = - 5 + i \left(- 15\right)$

$\textcolor{w h i t e}{{z}_{1} + {z}_{2}} = - 5 - 15 i$