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