# What is the trigonometric form of  (-10+3i) ?

Dec 30, 2015

$\left(- 10 + 3 i\right) = \sqrt{109} \left(C o s \left({\tan}^{-} 1 \left(- \frac{3}{10}\right)\right) + i \sin \left({\tan}^{-} 1 \left(- \frac{3}{10}\right)\right)\right)$

#### Explanation:

If $\left(a + i b\right)$ is a complex number, $u$ is its magnitude and $\alpha$ is its angle then $\left(a + i b\right)$ in trigonometric form is written as $u \left(\cos \alpha + i \sin \alpha\right)$.
Magnitude of a complex number $\left(a + i b\right)$ is given by$\sqrt{{a}^{2} + {b}^{2}}$ and its angle is given by ${\tan}^{-} 1 \left(\frac{b}{a}\right)$

Let $r$ be the magnitude of $\left(- 10 + 3 i\right)$ and $\theta$ be its angle.
Magnitude of $\left(- 10 + 3 i\right) = \sqrt{{\left(- 10\right)}^{2} + {3}^{2}} = \sqrt{100 + 9} = \sqrt{109} = r$
Angle of $\left(- 10 + 3 i\right) = T a {n}^{-} 1 \left(\frac{3}{-} 10\right) = {\tan}^{-} 1 \left(- \frac{3}{10}\right) = \theta$

$\implies \left(- 10 + 3 i\right) = r \left(C o s \theta + i \sin \theta\right)$
$\implies \left(- 10 + 3 i\right) = \sqrt{109} \left(C o s \left({\tan}^{-} 1 \left(- \frac{3}{10}\right)\right) + i \sin \left({\tan}^{-} 1 \left(- \frac{3}{10}\right)\right)\right)$