I know there must be something unmathematical in the following but I don't know where it is:

\begin{align} \sqrt{-1} &= i \\\\\ \frac1{\sqrt{-1}} &= \frac1i \\\\ \frac{\sqrt1}{\sqrt{-1}} &= \frac1i \\\\ \sqrt{\frac1{-1}} &= \frac1i \\\\ \sqrt{\frac{-1}1} &= \frac1i \\\\ \sqrt{-1} &= \frac1i \\\\ i &= \frac1i \\\\ i^2 &= 1 \\\\ -1 &= 1 \quad !!? \end{align}

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    $\begingroup$ When dealing with square roots, it is important to remember that each number apart from 0 naturally has 2 different roots. This means that you have to be very careful when dealing with non-positive numbers $\endgroup$
    – Casebash
    Jul 22, 2010 at 4:22
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    $\begingroup$ @Casebash: I believe it's fairly standard to take sqrt(x) to mean the principal square root function of x--that is, if x is a nonnegative real number, the nonnegative square root. Defining which root is the principal root of nonreal complex numbers can be a little trickier (some texts use the one with argument in [0,π) while other texts and many calculators use the one with argument in (-π/2,π/2] ). $\endgroup$
    – Isaac
    Jul 22, 2010 at 4:48
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    $\begingroup$ $i^2=(-i)^2$ doesn't imply $i=-i$. Simple but complex :-). $\endgroup$
    – copper.hat
    Jan 21, 2014 at 7:49
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    $\begingroup$ How $\sqrt{-1}=i$ if and only if $i^2=-1$? Or how can we conclude that $\sqrt{-9}=\pm 3i$? $\endgroup$
    – daulomb
    Jan 21, 2014 at 16:34
  • 6
    $\begingroup$ $\sqrt{-1}\neq \frac{1}{i} = -i$ $\endgroup$
    – AlvinL
    Mar 30, 2016 at 6:54

14 Answers 14


Between your third and fourth lines, you use $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$. This is only (guaranteed to be) true when $a\ge 0$ and $b>0$.

edit: As pointed out in the comments, what I meant was that the identity $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$ has domain $a\ge 0$ and $b>0$. Outside that domain, applying the identity is inappropriate, whether or not it "works."

In general (and this is the crux of most "fake" proofs involving square roots of negative numbers), $\sqrt{x}$ where $x$ is a negative real number ($x<0$) must first be rewritten as $i\sqrt{|x|}$ before any other algebraic manipulations can be applied (because the identities relating to manipulation of square roots [perhaps exponentiation with non-integer exponents in general] require nonnegative numbers).

This similar question, focused on $-1=i^2=(\sqrt{-1})^2=\sqrt{-1}\sqrt{-1}\overset{!}{=}\sqrt{-1\cdot-1}=\sqrt{1}=1$, is using the similar identity $\sqrt{a}\sqrt{b}=\sqrt{ab}$, which has domain $a\ge 0$ and $b\ge 0$, so applying it when $a=b=-1$ is invalid.

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    $\begingroup$ Well sqrt(-25) / sqrt(-1) = 5, so the explanation doesn't seem quite complete. $\endgroup$
    – Wilhelm
    Jul 22, 2010 at 4:21
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    $\begingroup$ Just because something is true for some cases, doesn't mean that it is true in other cases. The square root division law described by Isaac is the only defined way. sqrt(-25)/sqrt(-1) might = 5, but that's just a coincidence. There is only one rigorously defined way that keeps everything in order. $\endgroup$
    – Justin L.
    Jul 22, 2010 at 4:38
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    $\begingroup$ Thanks Justin and Himadri for clarifying my answer. Hopefully my edit will help as well. $\endgroup$
    – Isaac
    Jul 22, 2010 at 5:05
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    $\begingroup$ @workmad3: It's true that it works in both those cases, but it's misleading. It's comparable to saying that it's okay to cancel the 6s in 16/64 to get 1/4--it works, but for the wrong reasons. The identity sqrt(a) * sqrt(b) = sqrt(a * b) really does only work when a and b are nonnegative. $\endgroup$
    – Isaac
    Jul 22, 2010 at 13:33
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    $\begingroup$ @Isaac, I believe the key here originates from unit signs. Every number can be expressed in the form $N = e^{i \theta}*|N|$ where the $|N|$ gives the absolute value of the number and must strictly be greater than or equal to positive 0 while: $e^{i \theta}$ is the angle or orientation of the number in the complex number plane. The identity $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$ holds true whenever a,b both share the same orientation ($e^{i\theta}$) (ie exist on the same line from the origin). $\endgroup$ Jun 17, 2014 at 23:14

The simple reason this,

$$\sqrt{\frac{-1}{1}} =\frac{\sqrt{1}}{\sqrt{-1}}$$

is not valid is because of a branch cut that must be taken as a result of the square root function.

Using exponentials:

$$ \tag{1} e^{i\pi} = -1 $$ and thus, $$ \tag{2} e^{-i\pi} = \frac{1}{-1} = -1 $$

However, in the complex plane, these represent a semicircle rotation from $+1 \to -1$ in both anticlockwise and clockwise directions. The fact that both end up at $-1$ on the Real Axis simply conceals the path taken from $+1 \to -1$ in the complex plane.

Now if we think of both (1) and (2) they are two opposite semi-circles that make a circle in the complex plane of radius 1 around the origin: enter image description here

However, from complex analysis, we know that there must be a branch cut from $0 \to \infty$ somewhere on this circle. Thus, you are crossing a branch cut on the Riemann Surface by doing,

$$ \sqrt{\frac{-1}{1}} \to \frac{\sqrt{1}}{\sqrt{-1}} $$

To correctly, dodge the Branch Cut issues, we can use the exponential form ensuring that we simplify the exponentials inside the square root before taking the square root, $$ \sqrt{\frac{e^{i\pi}}{1}\frac{e^{-i\pi}}{e^{-i\pi}}} = \sqrt{\frac{e^{0}}{e^{-i\pi}}} = \sqrt{\frac{e^{i\pi}}{1}} = i $$ or similarly, $$ \sqrt{\frac{e^{-i\pi}}{1}\frac{e^{+i\pi}}{e^{+i\pi}}} = \sqrt{\frac{e^{0}}{e^{+i\pi}}} = \sqrt{\frac{e^{-i\pi}}{1}} = -i $$ The fact we get two different answers from using (1) and then using (2) to invert the terms inside the square root demonstrate the chaos we give ourselves when crossing Branch Cuts. Morale of Story: Don't Cross Branch Cuts as "Here be multivalued-Dragons" see bottom for picture.

You can also be really sneaky and do two full rotations, which if you look at the picture below, will put you where you started on the Riemann sheet.

See this image for a pictorial representation of the manifold which you are attempting to abuse (This is 3D version of the above and the branch cut can be seen where the sheet intersects itself). This picture also shows why we can take the Branch Cut anywhere as long as it is within one rotation as we can spin the shape in the vertical axis and it will remain the same.

enter image description here

A demonstration of what happens to an unprepared Math Student, (Here called 'Bob') who unknowingly crossed a Branch Cut without being prepared. enter image description here (credit also to @marios-bounakis for some exciting discussions about Euler identities)

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    $\begingroup$ This answer is flawed. It states implicitly that $\sqrt{1/-1}\ne \sqrt{-1/1}$. But in fact, $\sqrt{1/-1}=\sqrt{-1/1}$. The error is the OP was in writing $\sqrt{1}/\sqrt{-1}=\sqrt{1/-1}$. $\endgroup$
    – Mark Viola
    Feb 18, 2017 at 16:55
  • $\begingroup$ @MarkViola - incorporated your comment to make it clear that it is due to the square root and not simply the movement of signs from numerator to denominator within the square root. $\endgroup$ Mar 12, 2018 at 13:51
  • $\begingroup$ Do you have the source/reference for the 3D manifold, or at least the recipe for generating it? $\endgroup$
    – Cheng
    Sep 2, 2022 at 3:09
  • $\begingroup$ After dodging the Branch Cut Issue, we still get two different values, right? $\endgroup$
    – Cheng
    Sep 2, 2022 at 3:14

Isaac's answer is correct, but it can be hard to see if you don't have a strong knowledge of your laws. These problems are generally easy to solve if you examine it line by line and simplify both sides.

$$\begin{align*} \sqrt{-1} &= i & \mathrm{LHS}&=i, \mathrm{RHS}=i \\ 1/\sqrt{-1} &= 1/i & \mathrm{LHS}&=1/i=-i, \mathrm{RHS}=-i \\ \sqrt{1}/\sqrt{-1} &= 1/i & \mathrm{LHS}&=1/i=-i, \mathrm{RHS}=-i \\ \textstyle\sqrt{1/-1} &= 1/i & \mathrm{LHS}&=\sqrt{-1}=i, \mathrm{RHS}=-i \end{align*}$$

We can then see that the error must be assuming $\textstyle\sqrt{1}/\sqrt{-1}=\sqrt{1/-1}$.

  • $\begingroup$ Only problem I see with these, is that when one is shaky on the laws simplifying might be a problem. $\endgroup$
    – Wilhelm
    Jul 23, 2010 at 0:28
  • $\begingroup$ @Wilhelm: It works pretty well if you use the most certain laws first $\endgroup$
    – Casebash
    Jul 23, 2010 at 0:31
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    $\begingroup$ Why do you sometimes use $\hat{\imath}$ and sometimes use $i$? $\endgroup$
    – L. F.
    Apr 12, 2019 at 10:05

You use the rule $\sqrt{ab}=\sqrt a\sqrt b$, which indeed holds for $a,b\ge 0$, which is the first time in our life where we encounter a definition of square roots, namely that for non-negative real numbers $x$, $\sqrt x$ is the unique non-negative real number $y$ such that $y^2=x$): And clearly, if $u,v$ are nonnegative real numbers with $u^2=a$ and $v^2=b$, then $uv$ is a nonnegative real number with $(uv)^2=ab$ -- in other words, for $a\ge0$ and $b\ge 0$, we have $\sqrt{ab}=\sqrt a\sqrt b$.

However, without the assumption $a,b\ge0$, this rule no longer holds in general, i.e., for the very reason of leaving the realm where the above argument holds, the argument need no longer apply in every case. You stumbled across one such case: Conventionally, for negative real numbers $x$, we define $\sqrt x$ to be the unique imaginary number $y$ with positive imaginary part such that $y^2=x$. This way, $\sqrt{-1}=i$ (and not $-i$, even though also $(-i)^2=-1$). What happens to the rule $$\sqrt{ab}\stackrel?=\sqrt a\sqrt b$$ with this extension? If $u,v\in \Bbb R_{\ge0}\cup i\Bbb R_{>0}$ with $u^2=a$ and $v^2=b$, then certainly $(uv)^2=ab$, but we cannot always conclude that $uv\in \Bbb R_{\ge0}\cup i\Bbb R_{>0}$. In particular, if $u=v=i$, then $uv=-1\notin \Bbb R_{\ge0}\cup i\Bbb R_{>0}$.

Once we introduced imaginary numbers as values of square roots, we will also want to extend the definition of $\sqrt\,$ to all complex numbers as inputs. The problem persists: There is no way to pick one of the possible values of $\sqrt z$ in such a way for all $z\in\Bbb C$ that the multiplicative rule for square roots holds -- in fact, we just saw that it breaks much earlier.

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    $\begingroup$ and what happens to it when a or b is negative? Is there another formula or is this a dead end? $\endgroup$
    – Nick
    Jan 21, 2014 at 7:36
  • $\begingroup$ @Nick Simply separate the roots in terms of positive ones and i. Also, it is not incorrect to say that $\sqrt{\frac{1,-1}}=1/i=-i$, as $(-i)^{2}=-1=\frac{1,-1}}=1$, it's just that square roots have two possible values always. $\endgroup$ Apr 12, 2015 at 19:05
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    $\begingroup$ @HagenvonEitzen, adding "only" to the condition makes it necessary however it is a sufficient condition. Do you see the Isaac's answer contains a statement "This is only(guarenteed to be true) when ..."? This sentence show that it is a sufficient condition and not necessary. In particular, addition of the term "only" made your answer incorrect. $\endgroup$ May 10, 2015 at 11:35
  • $\begingroup$ @Hagen von Eitzen) I think your answer is wrong...Statement is TRUE when at least one of $a$ and $b$ is $\ge 0$. $\endgroup$
    – Empty
    Oct 5, 2015 at 19:00
  • $\begingroup$ As said in the comments, the phrasing makes this answer wrong. $\endgroup$ Mar 30, 2016 at 3:30

As others have mentioned, one approach is to be consistent about using principal roots, and make sure that the identities you're using are actually applicable for manipulating those principal roots. And your mistake there lies in equating the 3rd and 4th lines. But to me, that type of approach feels about as intuitive as memorizing a phone book.

Personally I find it much easier to think of square roots (or roots in general) as set-valued operators. Every number (other than zero) has two square roots, three cube roots, four fourth roots, etc (which in general can be complex).

Define the $n^{th}$ root to be the set-valued mapping: $\sqrt[n]x \mapsto \{ y : y^n = x \}$

Examples: \begin{gather} \sqrt{4} \mapsto \{ +2 , -2 \} \\\\ \sqrt{1} \mapsto \{ +1 , -1 \} \\\\ \sqrt{-1} \mapsto \{ -i, +i \} \\\\ \sqrt[3]{1} \mapsto \left\{ 1 , \tfrac{-1 + i \sqrt{3}}{2} , \tfrac{-1 - i\sqrt{3}}{2} \right\} \end{gather}

Then your question can be written as follows (treating multiplication and division as Kronecker products to see all the possible outcomes):

\begin{gather} \sqrt{-1} \mapsto \{ i, -i \} \\\\ \frac{1}{\sqrt{-1}} \mapsto \left\{ \frac{1}{i} , \frac{1}{-i} \right\} \\\\ \frac{1}{\sqrt{-1}} \mapsto \left\{ \frac{1}{i} , \frac{1}{-i} \right\} \\\\ \frac{\sqrt{1}}{\sqrt{-1}} \mapsto \frac{ \left\{ 1 , -1 \right\} }{ \left\{ i , -i \right\} } = \left\{ \frac{1}{i} , \frac{-1}{i} , \frac{1}{-i} , \frac{-1}{-i} \right\} = \left\{ \frac{1}{i} , -\frac{1}{i} \right\} = \left\{ -i , i \right\} \\\\ \sqrt{\frac{1}{-1}} = \sqrt{-1} \mapsto \left\{ i , -i \right\} \\\\ \sqrt{-1} \mapsto \left\{ i , -i \right\} \\\\ \left( \sqrt{-1} \right)^2 \mapsto \left\{ i^2 , \left(-i\right)^2 \right\} = \left\{ -1 , -1 \right\} = \{ -1 \} \end{gather} so \begin{gather} \left( \sqrt{-1} \right)^2 = -1. \end{gather} Also note that \begin{gather} \left( \frac{\sqrt{1}}{\sqrt{-1}} \right)^2 \mapsto \left\{ (-i)^2 , i^2 \right\} = \left\{ -1 , -1 \right\} = -1. \end{gather}


There is always a danger with dealing with roots of any kind, that one might not be dealing with the same number all the time. This comes in part from both $1^2=1$ and $(-1)^2=1$.

Writing these into an equation as $\sqrt{1}=1$ and $\sqrt{1}=-1$, gives a result that $a=-a$. Such might be true in some mantissa-space (mantissa here is a multiplication form of modulus: ie just as $a + bn = a \pmod{n}$, so would $a * n^b \operatorname{man} n$).

By taking square roots, one is effectively dealing in a potential mantissa-space where $+x=-x$, and some subtly is needed to distingiush the two. This is one of the reasons that $\sqrt{x} \ge 0$ is taken as convention.

The actual mistake in the calculations, is that $\sqrt{x}$ is let to vary by sign.


I think that there have been a number of approaches used to find $\frac{1}{i}$. Here may be a few to consider:

The positive powers of $i$ run in a cycle $(i, -1, -i, 1)$. Projecting this back into zero and the negative exponents, $i^0 = 1$ and $i^{-1}$ should be $-i$.

If we "rationalize" $\frac{1}{i}$ by multiplying both sides by $i$, then $\frac{1}{i} * \frac{i}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i$.


given $a,b \in \mathbb R $
Rule 1:
$\sqrt {a b} \iff \sqrt {a} \sqrt {b} $ is valid only if $a,b \geq0$
Similarly, $\sqrt { \frac{a}{b}} \iff \frac{\sqrt {a}}{ \sqrt {b}} $ is valid only if $a\geq 0,b>0$
Rule 2:
And,if $a>0$
Case 1: $\sqrt{a}$ will give only one positive value (more precise: you will get purely real complex number with positive real part) eg. $\sqrt{4}=2 $
Case 2: $\sqrt{-a} = i \sqrt{a}$ will give only one complex number with positive imaginary part. (more precise: you will get purely imaginary complex number with positive imaginary part) eg. $\sqrt{-4}= 2i $

So, you have done wrong in fourth line
Lets take a example, shows if you do not follows above rules you can create blunder. Take $a>0$
$\sqrt{-a} = i \sqrt{a} $
But if do not follow above rule you can do blunder as
$\sqrt{-a} = \sqrt{\frac{a}{-1}} $
$=>\sqrt{-a} = \frac{\sqrt{a}}{\sqrt{-1}} $
$=>\sqrt{-a} = -i \sqrt{a} $

All the above taking and discussion is done under the consideration that in general the notion of $\sqrt[n]{x}; x\in\mathbb C$ talking about only about principal root and principal root and it is unique.

But for all the roots of $\sqrt[n]{x}; x\in\mathbb C$ Some Mathematician generally uses $x^\frac{1}{n}; x\in\mathbb C$

For example:
$(4)^\frac{1}{2} = 2,-2$ { more precisely: $(4)^\frac{1}{2} \implies 2,-2$ but $2 \not\implies (4)^\frac{1}{2} $ or $(-2) \not\implies (4)^\frac{1}{2}$ }
Similarly, $(-1)^\frac{1}{2} = i,-i$ because $(-i)* (-i) = \sqrt{-1} = i *i$


Some of the confusion here is because there are two types of roots: roots and principal roots. The principal roots always return only one value. However symbolically they are almost never different. You can very, very rarely encounter $_+\sqrt{\;}\;$for the principal root. So confusion is often created because people confuse these two types of roots, as they are represented by exactly the same symbols. $$_+\sqrt1=1$$ $$\sqrt1=-1,+1$$ $$\sqrt{-1}=-i,+j$$ If we are doing something like this $$\sqrt{-1}=\frac{1}{i}$$ it means we have to take the correct roots; otherwise, we can have even this: $$\sqrt{1}=-1\;\ and \;\sqrt{1}=1$$ Hence, $1=-1\,.\;$ That's not the way to go. We have multiple roots when dealing with complex numbers, even square roots from real numbers have two answers. To avoid this, the principal root is used but it doesn't differ symbolically from just square root. People extremely rarely write $_+\sqrt{\;}\;$. Therefore there's a lot of confusion. The rules of extracting roots from complex numbers don't strictly follow the rules used with principal roots, so you may easily arrive at a wrong answer. So when we see, for example, the formula: $$\sqrt[n]{z}=\sqrt[n]{p}\, \bigg(\cos \frac{φ+2πk}{n}+i\sin\frac{φ+2πk}{n}\bigg)$$ We have to understand that what is meant is this $$\sqrt[n]{z}=\,_+\sqrt[n]{p}\, \bigg(\cos \frac{φ+2πk}{n}+i\sin{\frac{φ+2πk}{n}}\bigg)$$ Another example. We can't reduce a multiple root $\;\sqrt[nk]{z^k}\;$ to $\;\sqrt[n]{z}\;$ because the first one has $nk$ different root values and the second--only $n\,$.

Formulas such as $\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$ will work but not always. Generally, we can't use rules for principal roots for real numbers when we are dealing with complex numbers.

To recap:

First, $\sqrt{-1}=-i,+i\;\;$ ( NOT JUST $\;i\,$)

Second, when dealing with complex numbers, $\sqrt1=-1,+1\;\;$ ( NOT JUST $\;1\;$)

And third, $\sqrt{-1}\cdot\sqrt{-1}=1\;$ if we take $\;i,-i\;$ or $\;-i,i\;$ as roots, and $\sqrt{-1}\cdot\sqrt{-1}=-1\;$ if we take either $\;i,i$ or $\;-i,-i\;$ as roots. And even if we deal with real numbers and extract roots (not principal roots) we can still arrive at different values.

In very old books $\sqrt{-1}$ is sometimes used instead of $i$. It can only add to confusion.

Most crucial here: Roots must be distinguished from principal roots. When dealing with complex numbers we don't extract some principal root of them, but we have a set of different root values. So, almost every single line is wrong in the OP's 'proof'. Kevin Holt provided a very nice step by step illustration.


If $z$ and $w$ are two complex numbers, then it is not true in general $$ \sqrt{z}\sqrt{w}=\sqrt{zw} \tag{*}\label{*} \, . $$ The "rule" $\sqrt{\dfrac{z}{w}}=\dfrac{\sqrt{z}}{\sqrt{w}}$ also does not hold in general, as saying that $\sqrt{\dfrac{z}{w}}=\dfrac{\sqrt{z}}{\sqrt{w}}$ is equivalent to saying that $\sqrt{w}\sqrt{\dfrac{z}{w}}=\sqrt{z}$, which is a special case of $\eqref{*}$.

So why do these rules fail in the complex world, and what do we mean by the square root of a complex number in the first place? These are good questions, and the answers are more ... complex ... than one might expect.

When working in the real numbers, if $x$ is positive then the symbol "$\sqrt{x}$" refers to the positive number $a$ such that $a^2=x$. Although $2$ and $-2$ are both square roots of $4$, we designate $2$ as the "principal" root, and write $\sqrt{4}=2$. We adopt this notational convention so that $\sqrt{\cdot}$ may be regarded as a well-defined function from $[0,\infty)$ to $[0,\infty)$.

Things become trickier in the complex world, as there is no such thing as "positive" and "negative" complex numbers. (To be precise, there is no way of defining a total order $\leq$ on $\mathbb C$ so that it becomes an ordered field, and so any attempt to define the terms "positive" and "negative" will have limited success.) On what grounds can we say that $i$ is the principal square root of $-1$, then? Both $i$ and $-i$ are square roots of $-1$, but neither of them are positive. It is for this reason that some mathematicians would argue that the notation $i=\sqrt{-1}$ should be avoided entirely (see Ihf's answer to this question).

If we do insist upon defining the function $\sqrt{\cdot}$ in the complex world, then here is one possible approach. Every non-zero complex number $z$ can be uniquely written in the form $r\exp(i\theta)$, where $\theta \in (-\pi,\pi]$ (we say that $\theta$ is the principal argument of $z$) and $r>0$. If $z=r\exp(i\theta)$, then we could define $$ \sqrt{z} = \sqrt{r}\exp\left(\frac{i\theta}{2}\right) \, , $$ where $\sqrt{r}$ denotes the positive real root of $r$. This means that if $z=r_1\exp(i\theta_1)$ and $w=r_2\exp(i\theta_2)$ (where $\theta_1,\theta_2\in(-\pi,\pi]$ and $r_1,r_2>0$), then \begin{align} \sqrt{z}\sqrt{w} &= \sqrt{r_1}\exp\left(\frac{i\theta_1}{2}\right)\sqrt{r_2}\exp\left(\frac{i\theta_2}{2}\right) \\ &= \sqrt{r_1}\sqrt{r_2}\exp\left(\frac{i\theta_1}{2}\right)\exp\left(\frac{i\theta_2}{2}\right) \\ &= \sqrt{r_1r_2}\exp\left(\frac{i(\theta_1+\theta_2)}{2}\right) \, . \end{align} It might appear at first glance that the final line is the very definition of $\sqrt{zw}$, but this is not so. Notice that we defined $\sqrt{z}$ as $\sqrt{r}\exp\left(\frac{i\theta}{2}\right)$, we were careful to specify that $\theta$ is the principal argument of $z$. If $\theta_1+\theta_2\not\in(-\pi,\pi]$, then $\theta_1+\theta_2$ is not the principal argument of $zw$, and so we are not justified in writing $$ \sqrt{zw} = \sqrt{r_1r_2}\exp\left(\frac{i(\theta_1+\theta_2)}{2}\right) \, . $$ As it turns out, unless $\theta_1 + \theta_2$ is the principal argument of $zw$, the above equality does not hold, and so the "rule" $\sqrt{z}\sqrt{w}=\sqrt{zw}$ fails. In other words, for two non-zero complex numbers $z$ and $w$, the equality $\sqrt{z}\sqrt{w}=\sqrt{zw}$ holds if and only if the principal argument of $zw$ is the sum of the principal argument of $z$ and the principal argument of $w$. We can finally answer your question in detail. You wrote $$ \sqrt{\frac{1}{-1}} = \frac{\sqrt{1}}{\sqrt{-1}} \, , $$ which is equivalent to saying that $$ \sqrt{-1}\sqrt{\frac{1}{-1}} = \sqrt{1} \, . $$ or $$ \sqrt{-1}\sqrt{-1} = \sqrt{1} \, . $$ We can easily verify that this is incorrect because the $\text{LHS}=i^2=-1$, while the $\text{RHS}=1$. The deeper reason for why equality does not hold is because the principal argument of $-1$ is $\pi$, but the principal argument of $1$ is not $\pi+\pi=2\pi$.

What do we learn from this fake proof? Well, the $\sqrt{\cdot}$ function is much less well-behaved in the complex numbers compared to the real numbers, to the point where some mathematicians abstain from defining it entirely. More generally, we are reminded that blindly applying "rules" without checking whether we have satisfied the conditions to apply them inevitably leads to trouble.

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    $\begingroup$ This should be the accepted answer. $\endgroup$ Feb 22, 2023 at 19:45

$1, -1$ and $\sqrt 1$ are all real numbers.

$\sqrt{-1}$ is not a real number.

There is no reason to assume that the rules for using the $\sqrt{\phantom x}$ symbol apply to non real numbers. In fact, you have found out that they don't.

If by $\sqrt{-1 \times -1}$ you mean $\sqrt{(-1 \times -1)}$, then $\sqrt{-1 \times -1} = \sqrt{(-1 \times -1)} = \sqrt 1 = 1$.

Complex number are based on the creation of a new non real number, $i$. The "definition" of $i$ is $i^2 = i \cdot i = -1$.

Side comment: This is often expressed as $i$ is the square root of $-1$. In mathematics the word "the" usually implies that there is exactly one such thing. However, since it turns out that $(-i)(-i)= -1$, then $-1$ seems to have more than one square root. Hence is it incorrect to refer to $i$ as **the ** square root of $-1$. Good luck trying to correct that particular abuse of the language.

If we want to use the notation $i = \sqrt{-1}$ to indicate that $i$ is the square root of $-1$, then it follows, by definition , that

$$\sqrt{-1} \cdot \sqrt{-1} = i^2 = -1$$

and we already know that

$$\sqrt{(-1) (-1)} = \sqrt 1 = 1$$

We can prove that there is no real number whose square is $-1$. We had to invent a not-real number, $i$, to make one. It shouldn't come as too big of a surprise that $i$ does not behave entirely like a real number.


Guaranteeing that all $a$ and $b$ is larger than $0$, $\sqrt{a}\sqrt{b} = \sqrt{ab}$. This also can be used when only one is larger than zero. However, if those are all negetive, it comes out to be $-\sqrt{ab}$. This can be proven by figuring out the square roots of each two negative numbers. For instance, think about the numbers $-2$ and $-3$. You'll see that $\sqrt{-2}\sqrt{-3} = i\sqrt 2 i \sqrt 2 = i^2\sqrt 6 = -\sqrt 6.$ The fact that $\frac{\sqrt a}{\sqrt b}=-\sqrt{\frac{a}{b}}$ when a is positive and b is negative can be proven in the same way.


$\def\cmlt{\operatorname{cmlt}}$ $\def\csqr{\operatorname{csqr}}$

In this case $-1$ is seen as a complex number, and the square root function over complex numbers is multi-valued, and mostly $\sqrt{e^{2\pi i}}=-1$, whereas $\sqrt{e^{0}}=1$.

Define multiplication over $\mathbb{C}$ as:

$$\cmlt:\mathbb{C}\times\mathbb{C}\to\mathbb{C}$$ $$\cmlt(r_1e^{i \theta_1},r_2e^{i \theta_2})=r_1r_2e^{\theta_1+\theta_2}$$

We need to also specify the lower of $\theta_1$ and $\theta_2$ as the 'base' level for the multi-valued function, so that the multiplication is oriented in a positive fashion.

Define square root over $\mathbb{C}$ as:

$$\csqr:\mathbb{C}\to\mathbb{C}$$ $$\csqr(re^{i \theta})=\sqrt{r}e^{\frac{i\theta}{2}}$$

It is easy to see that, if $a,b$ are two complex numbers, then:


Now, $-1\equiv e^\frac{i\pi}{2}$ (via base levels defined above), and so both definitions return $-1$, this being $\sqrt{e^{2\pi i}}$.


what you are saying is similar to $1^2=(-1)^2$ so $1=-1$, math does not work that way


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