# How do you divide  (-4+9i)/(1-5i)  in trigonometric form?

Jul 17, 2017

$\frac{- 4 + 9 i}{1 - 5 i} = \frac{1}{26} \sqrt{2522} \left(\cos 167 - i \sin 167\right)$

#### Explanation:

$\frac{- 4 + 9 i}{1 - 5 i} = \frac{\left(- 4 + 9 i\right) \left(1 + 5 i\right)}{\left(1 - 5 i\right) \left(1 + 5 i\right)} = \frac{- 4 - 11 i - 45}{26} = \frac{- 11 i - 49}{26} = - \frac{49}{26} - \frac{11}{26} i$

Trigonometric form of a complex number:

$a + b i = r \left(\cos \vartheta + i \sin \vartheta\right)$, where $r = \sqrt{{a}^{2} + {b}^{2}}$ and $\vartheta = \arctan \left(\frac{b}{a}\right)$

$\sqrt{\frac{49}{26} ^ 2 + \frac{11}{26} ^ 2} = \frac{1}{26} \sqrt{2522}$

$\arctan \left(\frac{- \frac{11}{26}}{- \frac{49}{26}}\right) \approx - {167}^{\circ}$

$\therefore - \frac{49}{26} - \frac{11}{26} i = \frac{1}{26} \sqrt{2522} \left(\cos \left(- 167\right) + i \sin \left(- 167\right)\right)$

which can be rewritten as $\frac{1}{26} \sqrt{2522} \left(\cos 167 - i \sin 167\right)$