# How do you write the complex number in trigonometric form 5-12i?

Jun 6, 2017

In trigonometric form : $13 \left(\cos 292.62 + i \sin 292.62\right)$

#### Explanation:

$Z = a + i b$. Modulus: $| Z | = \sqrt{{a}^{2} + {b}^{2}}$; Argument:$\theta = {\tan}^{-} 1 \left(\frac{b}{a}\right)$ Trigonometrical form : $Z = | Z | \left(\cos \theta + i \sin \theta\right)$

$Z = 5 - 12 i$. Modulus $| Z | = \sqrt{{5}^{2} + {\left(- 12\right)}^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13$

Argument: $\tan \alpha = \frac{12}{5} = 2.4$. Z lies on fourth quadrant, $\alpha = {\tan}^{-} 1 \left(2.4\right) = {67.38}^{0} \therefore \theta = 360 - 67.38 = {292.62}^{0} \therefore Z = 13 \left(\cos 292.62 + i \sin 292.62\right)$

In trigonometric form expressed as $13 \left(\cos 292.62 + i \sin 292.62\right)$ [Ans]