Find $$$\sqrt[4]{1}$$$

Find $$$\sqrt[4]{1}$$$.

The polar form of $$$1$$$ is $$$\cos{\left(0 \right)} + i \sin{\left(0 \right)}$$$ (for steps, see polar form calculator).

According to the De Moivre's Formula, all $$$n$$$-th roots of a complex number $$$r \left(\cos{\left(\theta \right)} + i \sin{\left(\theta \right)}\right)$$$ are given by $$$r^{\frac{1}{n}} \left(\cos{\left(\frac{\theta + 2 \pi k}{n} \right)} + i \sin{\left(\frac{\theta + 2 \pi k}{n} \right)}\right)$$$, $$$k=\overline{0..n-1}$$$.

We have that $$$r = 1$$$, $$$\theta = 0$$$, and $$$n = 4$$$.

• $$$k = 0$$$: $$$\sqrt[4]{1} \left(\cos{\left(\frac{0 + 2\cdot \pi\cdot 0}{4} \right)} + i \sin{\left(\frac{0 + 2\cdot \pi\cdot 0}{4} \right)}\right) = \cos{\left(0 \right)} + i \sin{\left(0 \right)} = 1$$$
• $$$k = 1$$$: $$$\sqrt[4]{1} \left(\cos{\left(\frac{0 + 2\cdot \pi\cdot 1}{4} \right)} + i \sin{\left(\frac{0 + 2\cdot \pi\cdot 1}{4} \right)}\right) = \cos{\left(\frac{\pi}{2} \right)} + i \sin{\left(\frac{\pi}{2} \right)} = i$$$
• $$$k = 2$$$: $$$\sqrt[4]{1} \left(\cos{\left(\frac{0 + 2\cdot \pi\cdot 2}{4} \right)} + i \sin{\left(\frac{0 + 2\cdot \pi\cdot 2}{4} \right)}\right) = \cos{\left(\pi \right)} + i \sin{\left(\pi \right)} = -1$$$
• $$$k = 3$$$: $$$\sqrt[4]{1} \left(\cos{\left(\frac{0 + 2\cdot \pi\cdot 3}{4} \right)} + i \sin{\left(\frac{0 + 2\cdot \pi\cdot 3}{4} \right)}\right) = \cos{\left(\frac{3 \pi}{2} \right)} + i \sin{\left(\frac{3 \pi}{2} \right)} = - i$$$

$$$\sqrt[4]{1} = 1$$$A

$$$\sqrt[4]{1} = i$$$A

$$$\sqrt[4]{1} = -1$$$A

$$$\sqrt[4]{1} = - i$$$A