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