Polar form of $$$- 2 \sqrt{3} - 6 i$$$
Your Input
Find the polar form of $$$- 2 \sqrt{3} - 6 i$$$.
Solution
The standard form of the complex number is $$$- 2 \sqrt{3} - 6 i$$$.
For a complex number $$$a + b i$$$, the polar form is given by $$$r \left(\cos{\left(\theta \right)} + i \sin{\left(\theta \right)}\right)$$$, where $$$r = \sqrt{a^{2} + b^{2}}$$$ and $$$\theta = \operatorname{atan}{\left(\frac{b}{a} \right)}$$$.
We have that $$$a = - 2 \sqrt{3}$$$ and $$$b = -6$$$.
Thus, $$$r = \sqrt{\left(- 2 \sqrt{3}\right)^{2} + \left(-6\right)^{2}} = 4 \sqrt{3}$$$.
Also, $$$\theta = \operatorname{atan}{\left(\frac{-6}{- 2 \sqrt{3}} \right)} - \pi = - \frac{2 \pi}{3}$$$.
Therefore, $$$- 2 \sqrt{3} - 6 i = 4 \sqrt{3} \left(\cos{\left(- \frac{2 \pi}{3} \right)} + i \sin{\left(- \frac{2 \pi}{3} \right)}\right)$$$.
Answer
$$$- 2 \sqrt{3} - 6 i = 4 \sqrt{3} \left(\cos{\left(- \frac{2 \pi}{3} \right)} + i \sin{\left(- \frac{2 \pi}{3} \right)}\right) = 4 \sqrt{3} \left(\cos{\left(-120^{\circ} \right)} + i \sin{\left(-120^{\circ} \right)}\right)$$$A