# Question #191f5

Apr 6, 2017

$- 4 + 5 i \to \sqrt{41} \left(\cos \left({128.7}^{\circ}\right) + i \sin \left({128.7}^{\circ}\right)\right)$

#### Explanation:

To convert $a + b i \to r \left(\cos \left(\theta\right) + i \sin \left(\theta\right)\right)$

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

$r = \sqrt{- {4}^{2} + {5}^{2}}$

$r = \sqrt{16 + 25}$

$r = \sqrt{41}$

Because "a" is negative and "b" is positive we compensate for the second quadrant by adding ${180}^{\circ}$:

$\theta = {\tan}^{-} 1 \left(\frac{b}{a}\right) + {180}^{\circ}$

$\theta = {\tan}^{-} 1 \left(\frac{5}{-} 4\right) + {180}^{\circ}$

$\theta \approx {128.7}^{\circ}$

$- 4 + 5 i \to \sqrt{41} \left(\cos \left({128.7}^{\circ}\right) + i \sin \left({128.87}^{\circ}\right)\right)$