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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.

Pat Dolan
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3 Answers3

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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

Doug M
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  • Oh. I though I had accounted for both branches of the sqrt, albeit in a compact manner. Expanded: $+1 = +i^2 and -1 = -i^2$ – Pat Dolan Jan 27 '18 at 00:24
  • There is only one root. $x = \sqrt {i^4} \implies x = 1$ while $x^2 = i^4 \implies x = \pm 1$ – Doug M Jan 27 '18 at 01:00
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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 $(-1)^2=i^4=1.$

Michael Rozenberg
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    The radical symbol denotes the principal root. $\sqrt{i^4}$ equals $1$ and only $1$. – JMoravitz Jan 26 '18 at 23:48
  • I think you are total wrong. See please my definition. – Michael Rozenberg Jan 26 '18 at 23:50
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    If you wish to define it that way then your answer matches your definition. This is not the usual definition used by everyone else however. We generally like to define operations and such like this as functions from our set to the same set which your definition fails to be. – JMoravitz Jan 26 '18 at 23:53
  • In $\mathbb C$ it's not function. – Michael Rozenberg Jan 26 '18 at 23:54
  • Consider reading any analysis or complex analysis text which introduces and defines the radical symbol. Alternatively, if you trust wikipedia enough as a source, see this page and the passage "Roots are usually written using the radical symbol or radix with $\sqrt{x}$ denoting the principal square root of $x$, $\sqrt[3]{x}$ denoting the principal cube root..." – JMoravitz Jan 26 '18 at 23:56
  • There is another reasoning. But I think the essence does not change and it's important. Bye. – Michael Rozenberg Jan 27 '18 at 00:01
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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 $1=\sqrt{i^4}$.

"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 $1=\sqrt{(-1)\times(-1)}\color{red}{=}\sqrt{-1}\times\sqrt{-1}=i^2=-1$.

JMoravitz
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  • Yes, but when we take principal roots we want what is in the parentheses to be between -pi and pi, in this case for I it would be pi/2 and for i^4 would be zero. – JMoravitz Feb 08 '18 at 17:37
  • 4 times pi/2 is 2pi, and before we take the square root, we rewrite the exponent so the coefficient of i in the exponent is between pi and -pi, which would in this case be zero. Half of zero is still zero – JMoravitz Feb 08 '18 at 17:45
  • In the earlier duplicate of this question Unem Chan ingeniously avoided the inherent confusion in taking the square root of $i^4$ by deducing the same contradiction based on Euler's identity. $$i=e^{\left(\frac{\pi}2+2k\pi\right)i}$$ $$i^4=e^{4\left(\frac{\pi}2+2k\pi\right)i}=e^{\left(2\pi+8k\pi\right)i}$$ $$\sqrt{i^4}=e^{\frac12\left(2\pi+8k\pi\right)i}=e^{\left(\pi+4k\pi\right)i}=-1$$ If we consider only the principle roots of both sides of his derivation then it is obvious that +1 cannot be simultaneously derived on both sides of the derivation. – Pat Dolan Feb 08 '18 at 18:00
  • -1 can be derived on left side of Chan. But only at the expense of permitting the inherent confusion involved in taking the square root of $i^4$ In that case we are delivered back into the clutches of my earlier derivation. – Pat Dolan Feb 08 '18 at 18:03
  • You still seem to be missing the point. Taking the square root of a number is not the same thing as reducing the exponent by half! Taking the square root is first rewriting the exponent into standard form with an argument in the correct range and only after this had been done do you reduce the exponent by half. The standard Argument of i^4 is zero. There is no contradiction. Sqrt(i^4)=1 – JMoravitz Feb 08 '18 at 18:06
  • Absolutely J. I agree. We absolutely restrict the value of $i^4$ to +1 in strict accordance with what you type. Therefore +1 IS THE PRINCIPLE ROOT in question on the left of Chan. But the problem for you is on the other side of Chan, which evaluates to -1 for all integer values of k. What am I missing here? – Pat Dolan Feb 08 '18 at 18:18
  • (2pi+8kpi)i=(0+2(4k+1)pi)i. The leftmost number in the exponent's coefficient of i should be in the correct range. Phrased differently, mod the exponent's coefficient of i out by 2pi. – JMoravitz Feb 08 '18 at 18:21
  • Again, I agree. But now perform the next step in Chan by using the law of powers to get (0+4K+1)pi. – Pat Dolan Feb 08 '18 at 18:34
  • Trying again to phrase it more carefully. Given an exponent of the form e^(ai) we first rewrite a=b+2pi c with b a real number between 0 and 2pi, allowed to be zero, not allowed to be 2pi, and c an integer, and before applying the square root to e^(ai) we write it as e^(bi). We do not include any even multiple of 2pi i in the exponent other than zero. We would as such have i^4=e^((2pi+8kpi)i)=e^((0+2pi(4k+1))i)=e^0. And then when we take the square root to this standardized form of writing i^4 we get 1 like we should. – JMoravitz Feb 08 '18 at 18:37
  • But does your expansion really accomplish your purpose? Let's expand again: e^((0+2pi(4k+1))i = e^(0i) x e^(2pi4k)i x e^(2pi)i. Now taking the sqrt of your expression in this new form we get 1 x e^(pi4k)i x e^(pi)i which always evaluates to -1. And isn't this the heart of the contradiction? There is no unique way to validly represent the expression. Hence no unique value to agree on. All we have are our personal preferences. – Pat Dolan Feb 08 '18 at 22:56
  • @patdolan There is in fact a unique way. One and only one way. That is the point. You made an error in taking the square root of e^(2pi i). The square root of e^(2pi i) is the square root of e^0 is 1. 2pi is outside the allowable range and so needed to be discarded. Given an imaginary exponent, there is one and only one result modulo 2pi i. – JMoravitz Feb 08 '18 at 23:27
  • The point is that we can and do have a predetermined method to have a unique standard representation for any number. There is and never has been a contradiction. – JMoravitz Feb 09 '18 at 00:03
  • If, as you state above, that the square root of e^(2pi i) is the square root e^0 then squaring both sides leads to the conclusion that e^(2pi i) = e^0 = 1 which further leads to ln(e^2pi i) = ln(e^0) = ln(1) which leads to the contradiction 2pi i = 0 or i = 0. – Pat Dolan Feb 09 '18 at 08:34
  • Incorrect again. There is a complex version of the natural logarithm which does similarly to the above and utilizes the standard form, notated as Ln or Log. See https://en.m.wikipedia.org/wiki/Complex_logarithm and the section discussing principal values. – JMoravitz Feb 09 '18 at 11:27
  • lim[x->2pi] of cos(x/2) - isin(x/2) != lim[x->0] of cos(x/2) - isin(x/2) – Pat Dolan Feb 12 '18 at 00:04
  • @PatDolan which is one of the unfortunate things about branch cuts. Things aren't necessarily going to be continuous like they were in the purely real case. That is one of the concessions that is made in order to avoid the "functions" from being multivalued (and thus not functions at all), and instead have a single value associated to each input (regardless of representation) and behave as nicely as they can. – JMoravitz Feb 12 '18 at 00:13
  • How you still continue to try to insist that what I'm saying is wrong or flawed somehow despite numerous citations and explanations to the contrary and still try to find supposed flaws in the argument which are not in fact flaws at all, just merely mistaken calculations on your part or mistaken assumptions on how limits should work for complex functions, it is quite frustrating. It is like you want me to be wrong and aren't willing to give this possibility which apparently conflicts with your worldview more than a second glance. – JMoravitz Feb 12 '18 at 00:20