Polar form of $$$-8$$$

Find the polar form of $$$-8$$$.

The standard form of the complex number is $$$-8$$$.

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 $$$a = -8$$$ and $$$b = 0$$$.

Thus, $$$r = \sqrt{\left(-8\right)^{2} + 0^{2}} = 8$$$.

Also, $$$\theta = \operatorname{atan}{\left(\frac{0}{-8} \right)} + \pi = \pi$$$.

Therefore, $$$-8 = 8 \left(\cos{\left(\pi \right)} + i \sin{\left(\pi \right)}\right)$$$.

$$$-8 = 8 \left(\cos{\left(\pi \right)} + i \sin{\left(\pi \right)}\right) = 8 \left(\cos{\left(180^{\circ} \right)} + i \sin{\left(180^{\circ} \right)}\right)$$$A