# Why $\sqrt{-1 \times -1} \neq \sqrt{-1}^2$? [duplicate]

We know $$i^2=-1$$then why does this happen? $$i^2 = \sqrt{-1}\times\sqrt{-1}$$ $$=\sqrt{-1\times-1}$$ $$=\sqrt{1}$$ $$= 1$$

EDIT: I see this has been dealt with before but at least with this answer I'm not making the fundamental mistake of assuming an incorrect definition of $i^2$.

• Why is $\sqrt{-1} \cdot \sqrt{-1} = \sqrt{-1 \cdot -1}$? We know that $\sqrt{ab} = \sqrt{a} \sqrt{b}$ but this is only true for $a,b \geq 0$... Commented Jul 3, 2011 at 9:17
• Duplicate of math.stackexchange.com/q/438/7850 Commented Jul 3, 2011 at 11:49
• @TheChaz: while I agree that deep down the two questions are duplicates, I also think that a person who understands that the two questions are the same would not be having this confusion. So closing this as a duplicate may not actually help directly. (This is one of those issues having to do with abstract duplicates, ideally we should have a master question dealing precisely with problems about incorrect algebraic manipulations of $\sqrt{-1}$.) Commented Jul 3, 2011 at 12:29
• In every math textbook! If there is a math textbook that says $\sqrt{ab} = \sqrt{a}\,\sqrt{b}$ without providing restrictions on $a,b$, then you should write to the author immediately! Commented Jul 3, 2011 at 12:43
• $i^2=-1$ is the definition of $i$. Commented Feb 8, 2013 at 18:00

From $i^2=-1$ you cannot conclude that $i=\sqrt{-1}$, just like from $(-2)^2 = 4$ you cannot conclude that $-2=\sqrt 4$. The symbol $\sqrt a$ is by definition the positive square root of $a$ and is only defined for $a\ge0$.

It is said that even Euler got confused with $\sqrt{ab} = \sqrt{a}\,\sqrt{b}$. Or did he? See Euler's "mistake''? The radical product rule in historical perspective (Amer. Math. Monthly 114 (2007), no. 4, 273–285).

• If I cannot conclude that $i=\sqrt{-1}$ then what good is it? Wouldn't that mean I couldn't use $i$ for what it's intended? Namely $\sqrt{-5} = \sqrt{-1\times 5} = i\sqrt{5}$ ?
– Greg
Commented Jul 4, 2011 at 13:04
• @Greg, then only property of $i$ that you need is that $i^2=-1$. The symbol $\sqrt{-1}$ is abuse of notation. As for $\sqrt{-5}$, you get $(i\sqrt{5})^2= (-1)\cdot 5 = -5 =$, as you expect.
– lhf
Commented Jul 4, 2011 at 15:46
• There's nothing wrong with defining(!) $\sqrt{-1}=i$, just as there's nothing wrong with defining $\sqrt{4}=2$ despite the fact that from $x^2=4$ you cannot conclude $x=2$. The difference to real numbers is that you cannot define the square root function so that it is continuous on all of $\mathbb C$. And the calculation in the question is just incorrectly applying power laws; you don't even need complex numbers to do something similar: $-1 = (-1)^1 = (-1)^{2\cdot1/2} = ((-1)^2)^{1/2} = 1^{1/2} = 1$ Commented Jul 10, 2013 at 12:40
• So the power laws like $a^n*b^n=(a*b)^n$ are only applicable, when $a,b > 0$?
– Mike
Commented Dec 11, 2019 at 23:17

Any non zero number has two distinct square roots. There's an algebraic statement which is always true : "a square root of $a$ times a square root of of $b$ equals a square root of $ab$", but this does not tell you which square root of $ab$ you get.

Now if $a$ and $b$ are positive, then the positive square root of $a$ (denoted $\sqrt{a}$) times the positive square root of $b$ (denoted $\sqrt{b}$) is a positive number. Thus, it's the positive square root of $ab$ (denoted $\sqrt{ab}$). Which yields

$$\forall a,b \ge 0, \ \sqrt{a} \sqrt{b} = \sqrt{ab}$$

In your calculation, because $i$ is a square root of $-1$, then $i^2$ is indeed a square root of $1$, but not the positive one.

The problem is this: the (polar) representation of a complex number depends on your choice of branch. Once you choose a branch of the square root, you cannot simultaneously represent $$1$$, and $$-1$$, because they are $$\pi$$ apart, so that $$-1$$ and $$1$$ will necessarily be in different branches of $$z^{1/2}$$. The choice of square root you make (if you want it to be well-defined) depends on the choice of branch you are working with. You are trying to combine numbers that live in different branches; it is as if you add $$1+1$$ and get $$1-i\pi$$, since 1 can also be represented as $$i\pi$$ ($$~1$$ can actually be represented as $$ik\pi$$, but when you operate, you are supposed to stay within a branch, and, in your case, you are not$$~$$).

In complex numbers, from the perspective of polar representations, when you multiply $$z_1 \dot z_2$$, you multiply the respective lengths, and add the respective angles (but you have to make up for the fact that the sum of the angles may be larger than $$2\pi$$ (or whatever argument-system you are working with. So in this sense, when you multiply $$i$$ by itself, you multiply the length of $$i$$ by itself, and add the argument to itself; i.e., you double the argument. In complex variables, you have many possible polar representations for a given number; specifically, given $$z=re^{it}$$, then $$z=re^{i(t+2\pi)}$$ is also a valid representation.

You must then choose a specific representation for your $$z$$, specifically; you will need to specify the range of the argument you will be working with. So, say you work with the "standard" range of $$[0,2\pi)$$. Then the expression for $$i$$ (me) is as $$i=1e^{i\pi/2}$$, so that $$i^2= (1)(1).e^{i(\pi/2+\pi/2)}=e^{i\pi}=-1$$. But the way backwards from multiplying to taking roots is more complicated if your base is non-negative, and/or your exponent has non-zero imaginary part. When this last is the case, you define:

$$z^{{1}/{2}} \:=e^{{\Large{\log(z)}}},$$ where we define:

$$\log(z):=\ln |z|+i \arg(z)$$

(This choice of definition has to see in part with wanting to have the complex log agree with the standard real $$\log$$--though this agreement is possible only for one choice of "branch", as we will see.) But because of the infinitely-many possible choicesfor the argument of a number, the $$\log(z)$$ itself--defined locally as the inverse of $$e^z$$ is somewhat-ambiguously-defined, since $$e^z$$ does not have a global inverse (since it is not $$1-1$$, for one thing, but $$e^z$$ does have local inverses, e.g., by using the inverse function theorem). So when we mention $$\log$$, we are referring just to one of (infinitely-) many possible local inverses of $$e^z$$ .Each possible local inverse to $$e^z$$ is called a "branch" of the $$\log$$. So once we choose a branch for the $$\log$$, which is a choice of an open set (technically, it is half-open) of width $$2\pi$$ from which we will choose the argument we will use. So, say we choose the standard branch $$(0,2\pi)$$, which we call Log(z). We then define :

$$z^{1/2}:=e^{\large{\operatorname{Log}(z)/2}}$$

But, in this branch , $$(-1)^{1/2}$$ is not even defined, because the argument for $$(-1)$$ is $$0$$ , which is outside of the allowable values $$(0,2\pi)$$. So, in this sense, the expression $$(-1)^{1/2}$$ is not well-defined, i.e, does not really make sense. ($$~$$Note that this particular branch of $$\log$$ reduces to the standard one when you select an argument of $$t=0~$$).

• Maybe I overdid it, trying to understand the log better. Could someone help with the spacing? I am saturated on formatting, sorry.
– gary
Commented Jul 3, 2011 at 10:58
• Help exactly with what spacing in your answer? Commented Jul 4, 2011 at 2:34
• Well, I try to space out the paragraphs, but they do not come out with the spacing of the original; specifically, in spreading the text out some more, and in separating paragraphs.
– gary
Commented Jul 5, 2011 at 17:26
• @ gary: Alright got it. Commented Jul 7, 2011 at 13:48
• Another downvote without a constructive comment. Is it personal? If there is something specific you do not like, it would be nice to hear; I may be able to learn from it. Still, it is up to you, whomever downvoted me without an explanation.
– gary
Commented Jul 12, 2011 at 1:12

Following Willie Wong's suggestion, I will put the comments I made before into an answer. The problem with this argument is that the formula $\sqrt{ab} = \sqrt{a} \sqrt{b}$ is valid for $a,b \geq 0$. In particular, it is not valid in the case when $a=b=-1$ so that the step $\sqrt{-1}\sqrt{-1} = \sqrt{(-1)(-1)}$ is incorrect.

Don't take me the wrong way, all of you are right from a certain point of view.

Let me show you the following. Let $a$ and $b$ be complex numbers. If $\sqrt a$ is a square root of $a$ and $\sqrt b$ is a square root of $b$, then $(\sqrt a\sqrt b)^2 = (\sqrt a)^2 (\sqrt b)^2 = ab$ by assumption, so indeed $\sqrt a\sqrt b$ is a square root of $ab$. Notice that it is some square root; there is no canonical choice over $\mathbb{C}$!

From this point of view, your arguments work until $i^2 = \sqrt1$. All you've shown is that $-1 = i^2$ is a square root of 1 and no one will disagree with that fact.

So what I'm trying to say is, if you want to write something like $\sqrt a\sqrt b = \sqrt{ab}$, which isn't the worst idea, you can't write something like $\sqrt 1 = 1$.

@Greg: $i^2=-1$ is a matter of definition. It was introduced to handle a negative sign under the radical when solving polynomial equations. The best example is $x^2 +1 =0$, which yields $x = \pm\sqrt{-1}$. Using your manipulation above would yield $x^2 = 1$ which clearly does not solve the equation.

You may only apply $\sqrt{a} \sqrt{b} = \sqrt {ab}$ when $a,b \ge0$.

There are many ways to show that your second equality is incorrect. Just for the sake of argument and a refreshing change of pace, suppose your point is true. That is, suppose $i^{2} = 1$. Then \begin{align} (x + i)(x-i) = x^2 - i^{2} = x^{2} - 1. \end{align} Use the Descartes Rule of signs to derive a contradiction. Hint: $i$ is not real. Can you finish the line of reasoning and derive an absurdity?

• Isn't a quicker absurdity $-1 = i^2 = 1$? Commented Jul 3, 2011 at 9:33
• Hmm. if $1+i=0$ and $1-i=0$ then you can set them equal to each other. So $1+i=1-i$ which means $i=-i$. But I'm not sure why you conclude $(1+i)(1-i)=0$.
– Greg
Commented Jul 3, 2011 at 9:33
• @user02138: The OP is assuming $i$ is the square root of $-1$ in the derivation. If you're assuming nothing about $i$ then there's no contradiction, since for all you know $i$ could equal 1. Commented Jul 3, 2011 at 9:46

In general,

$\forall a,b \in \mathbb{R} , ab\geq 0, (\sqrt{ab} =\sqrt{ |{a}| }\sqrt{|b|})$