# How do you add (8-8i)+(-7-i) in trigonometric form?

Feb 15, 2017

1-9i = sqrt(82)(cos(arctan(-9)) + i sin ((arctan (-9)))

#### Explanation:

Add them in rectangular form $\left(a + b i\right)$ and then convert to trigonometric form:

1. Combine the like terms: $\left(8 + - 7\right) + \left(- 8 i + - i\right) = 1 - 9 i$
2. $r = \sqrt{{1}^{2} + {\left(- 9\right)}^{2}} = \sqrt{82}$
3. $\theta = \arctan \left(\frac{b}{a}\right) = \arctan \left(- \frac{9}{1}\right) = \arctan \left(- 9\right)$
4. $z = r \left(\cos \theta + i \sin \theta\right)$
= sqrt(82)(cos(arctan(-9)) + i sin ((arctan (-9)))