# What is the trigonometric form of -8-i?

Feb 23, 2018

$- \left(8 + i\right) \approx - \sqrt{58} \left(\cos \left(0.12\right) + i \sin \left(0.12\right)\right)$

#### Explanation:

$- 8 - i = - \left(8 + i\right)$

For a given complex number, $z = a + b i$, $z = r \left(\cos \theta + i \sin \theta\right)$

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

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

Let's deal with $8 + i$

$z = 8 + i = r \left(\cos \theta + i \sin \theta\right)$

$r = \sqrt{{8}^{2} + {1}^{2}} = \sqrt{65}$

$\theta = {\tan}^{-} 1 \left(\frac{1}{8}\right) \approx {0.12}^{c}$

$- \left(8 + i\right) \approx - \sqrt{58} \left(\cos \left(0.12\right) + i \sin \left(0.12\right)\right)$