What are the 4th degree roots of $1$?

Question

So my question is the fourth root of $1 = i$?

Where $i^4 = 1$? or if it is still $1$ nonetheless?

Answer 1

Remember $i$ is an imaginary number defined as $$\color{red}{i=\sqrt{-1}}\ \ \ $$$$\text{or } i^2=-1$$ The forth roots of $1$ are given as $$1^{1/4}=(\cos0+i\sin 0)^{1/4}$$ $$=\left(\cos(2k\pi)+i\sin(2k\pi)\right)^{1/4}$$ $$=\left(\cos\left(\frac{k\pi}{2}\right)+i\sin\left(\frac{k\pi}{2}\right)\right)$$ where, $k=0, 1, 2, 3$ hence, all forth roots of $1$ are given as $$1, \ \ i, \ \ -1, \ \ -i$$

hence, $i$ is one of the fourth roots of $1$ (unity)

and, one should have $$i^4=(i^2)(i^2)=(-1)(-1)=1$$

Answer 2

There are four fourth roots of $1$ in $\mathbb C$. Those are $\pm1$ and $\pm i$. Two of them are "primitive" fourth roots of $1$, meaning you don't get $1$ if you raise them to any lower power than $4$. You can raise $-1$ to a lower power than $4$, namely $2$, and get $1$, and you can raise $1$ to a lower power than $4$, namely $1$, and get $1$.

However, by convention, the notation $\sqrt[4]{1}$ means the positive fourth root of $1$.

Answer 3

It's not necessarily correct to talk about the fourth root of a number, in the same way that it's not necessarily correct to talk about the second (square) root of a number. In the latter case, if $x^2 = y$, then $(-x)^2 = y$, so both $x$ and $-x$ are second roots of $x$. We'll often talk about "the" square root of $x$ as the output $\sqrt{x}$ of the "square root function" by picking a "handy" second root of $x$ for every $x$ we define it on. For example, I could've defined \begin{align*} g(x) & = \begin{cases} \sqrt{x} & x \geq 0, x \in \mathbb{Q} \\ - \sqrt{x} & x \geq 0, x \not \in \mathbb{Q}, \end{cases} \end{align*} and $g$ would still spit out a square root of every input. In the case of $\mathbb{C}$, every non-zero complex number has $n$ unique $n^{\textrm{th}}$ roots. So though $i$ is a fourth root of $1$, it is not correct to call it the fourth root of $1$, as $ - i, \pm 1$ are also fourth roots of $1$.

Answer 4

In order to solve $\sqrt[n]{1}$:

• Draw the unit circle
• Draw the first solution, which is obviously $1+0i=\cos(0)+\sin(0)i$
• Repeat $n-1$ times: find the next solution by rotating the previous solution $\frac{2\pi}{n}$ radians

For example, $\sqrt[5]{1}$:

• $\cos(0)+\sin(0)i$
• $\cos(\frac{2\pi}{5})+\sin(\frac{2\pi}{5})i$
• $\cos(\frac{4\pi}{5})+\sin(\frac{4\pi}{5})i$
• $\cos(\frac{6\pi}{5})+\sin(\frac{6\pi}{5})i$
• $\cos(\frac{8\pi}{5})+\sin(\frac{8\pi}{5})i$