Does $\sqrt{i^4}$ imply +/-1 = -/+1?

Question

I know that the minus sign of -r, where r is real, can never be moved under a radical sign. But there are no minus signs present when the sqrt of $i^4$ is taken in the following manner

$1 = i^4$

$\sqrt{1} = \sqrt{i^4}$

$+/-1 = +/-i^2$

$+/-1 = -/+1$

I realize it might be argued that i also can never be moved under the radical sign. But $\sqrt{i}$ does have an open form solution which is $+/-[(1+i)/\sqrt2]$. Repeating the above calculation with this open form I find that the associative law of multiplication no longer holds and all sorts of contradictions ensue. It might also be argued that $i^4$ factors into -1 x -1 and therefore cannot be placed under the radical sign. But that is also true of any positive real number. Any ideas for a consistent solution to the above derivation? Thanks.

Answer 1

In $\mathbb C$ we have by definition: $\sqrt{z}=\{w|w^2=z\}$.

Thus,

$$\sqrt{i^4}=\{1,-1\}$$ because $1^2=i^4=1$ and $(-1)^2=i^4=1.$

Answer 2

When you introduce the radical ($\sqrt{}$), you are referring to the principal root of the object under the radical. There is only one principal root. And, you should be clear whether you are working in a complex number system or a real number system.

i.e. $\sqrt[3]{-1} = -1$ in the real numbers and $\frac 12 + \frac {\sqrt 3}{2} i$ in the complex numbers.

$\sqrt {i^4} = \sqrt 1 = 1$

While $x^2 = 1$ has $2$ solutions

Answer 3

The radical symbol $\sqrt{~~}$ is used to very specifically denote the principal root of a number. See $n$'th roots on wikipedia. The definition is not perfect since $f(x)=\sqrt{x}$ is not a continuous function on $\Bbb C$, however despite this it is the standard definition.

While $x^2=4$ may have two solutions, both $2$ and $-2$, $\sqrt{x}$ has only a single outcome. For example, $\sqrt{4}=2$ and only $2$, not $-2$. This is because although $-2$ is a square root of $4$, it is not the square root (which is intended to be interpreted as the principal root) of $4$.

$x^2=y^2\implies x=\pm y$ is a way of saying that $x^2=y^2$ implies that one of the two things will be true: that $x=y$ or that $x=-y$. It does not imply that they are both true simultaneously and it does not imply that $\sqrt{x^2}=\{x,-x\}$. Rather, $\sqrt{x^2}=|x|$ for real numbers, and more generally $\sqrt{x^2}=\begin{cases}x&\text{if }0\leq Arg(x)<\pi\\-x&\text{if }\pi\leq Arg(x)<2\pi\end{cases}$ where $Arg(x)$ denotes the principal argument of a complex number (in the case of real positive numbers this is $0$, and real negative numbers this is $\pi$).

As for the specifics of your question, yes, $1=i^4$. Yes, $\sqrt{1}=\sqrt{i^4}$. No, $1\neq i^2$

One has $1=\sqrt{1}$ and $1=\sqrt{i^4}$.

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"I find that the associative law of multiplication no longer holds and all sorts of contradictions ensue"

You may be operating under the assumption that various laws you learned for positive real numbers work for all numbers which is not the case. Examples of these would be:

• $\sqrt{a}\times \sqrt{b}=\sqrt{a\times b}$
• $(a^b)^c = (a^c)^b$
• $\sqrt{a^{2k}}=a^k$

Each of these can be false when $a$ is not a positive real number. Forgetting this can lead to contradictions such as $1=\sqrt{1}=\sqrt{i^4}\color{red}{=}i^2=-1$ or $1=\sqrt{(-1)\times(-1)}\color{red}{=}\sqrt{-1}\times\sqrt{-1}=i^2=-1$.