Polar form of $$$- \frac{1}{2} - \frac{\sqrt{3} i}{2}$$$
Your Input
Find the polar form of $$$- \frac{1}{2} - \frac{\sqrt{3} i}{2}$$$.
Solution
The standard form of the complex number is $$$- \frac{1}{2} - \frac{\sqrt{3} i}{2}$$$.
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 $$$a = - \frac{1}{2}$$$ and $$$b = - \frac{\sqrt{3}}{2}$$$.
Thus, $$$r = \sqrt{\left(- \frac{1}{2}\right)^{2} + \left(- \frac{\sqrt{3}}{2}\right)^{2}} = 1$$$.
Also, $$$\theta = \operatorname{atan}{\left(\frac{- \frac{\sqrt{3}}{2}}{- \frac{1}{2}} \right)} - \pi = - \frac{2 \pi}{3}$$$.
Therefore, $$$- \frac{1}{2} - \frac{\sqrt{3} i}{2} = \cos{\left(- \frac{2 \pi}{3} \right)} + i \sin{\left(- \frac{2 \pi}{3} \right)}$$$.
Answer
$$$- \frac{1}{2} - \frac{\sqrt{3} i}{2} = \cos{\left(- \frac{2 \pi}{3} \right)} + i \sin{\left(- \frac{2 \pi}{3} \right)} = \cos{\left(-120^{\circ} \right)} + i \sin{\left(-120^{\circ} \right)}$$$A