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Why is $1=\sqrt{(-1)(-1)}=\sqrt{i^2\cdot i^2}=\sqrt{i^4}=\sqrt{\left(e^{i\frac{\pi}{2}}\right)^4}=\sqrt{e^{i2\pi}}=e^{i\pi}=-1$ wrong?

my friend showed me this "proof" - what's the first step that is not allowed? Isn't $1=\sqrt{(-1)(-1)}$ wrong?

José Carlos Santos
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Analysis
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    @MartinR It may be a duplicate, but not of that question. The equality $\sqrt{ab}=\sqrt a\sqrt b$ is never used here. – José Carlos Santos Aug 08 '19 at 21:09
  • Each non-zero complex number has 2 square roots which are opposite of one another. The notation $\sqrt{x}$ is reserved for the positive square root of a positive real number $x$. (There is no such thing as $\sqrt{-1}$ despite of what some people write, but there are two different square roots of $-1$: $\pm i$, or $\mp i$ if you prefer.) – Alexey Aug 08 '19 at 21:10
  • @Alexey In this proof, the expression $\sqrt a$ is always used when $a\in[0,\infty)$. – José Carlos Santos Aug 08 '19 at 21:11
  • @JoséCarlosSantos, yes, and the value cannot be negative. – Alexey Aug 08 '19 at 21:12
  • @ParParabolicAlcoholic, what could be wrong with $1 =\sqrt{1}$?? – Alexey Aug 08 '19 at 21:14
  • @JoséCarlosSantos: What about this one https://math.stackexchange.com/questions/912670/when-does-eab-eab-hold ? – Martin R Aug 08 '19 at 21:14
  • @MartinR That's a better choice, but I would not vote to close it as a duplicate of that one either. Note that that question doesn't even mention square roots. – José Carlos Santos Aug 08 '19 at 21:45

4 Answers4

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The only incorrect statement is that $$\sqrt{e^{i2\pi}}=e^{i\pi}$$ Because we have that $$\sqrt{x^2}=|x|$$ for real $x$. So we would have the answer as $$\sqrt{e^{i2\pi}}=|e^{i\pi}|=|-1|=1$$

Peter Foreman
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There is nothing wrong with the equality $1=\sqrt{(-1)(-1)}$. The problem lies in the equality $\sqrt{e^{2\pi i}}=e^{\pi i}$, because $e^{2\pi i}=1$, and therefore $\sqrt{e^{2\pi i}}=1\neq e^{\pi i}$.

José Carlos Santos
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    Slightly more general: we don't necessarily have $(z^a)^b=z^{ab}$ for $z\in\mathbb C$ unless additional restrictions are put on $a,b$. – Clayton Aug 08 '19 at 21:12
  • @Clayton, $z^a$ is not even defined for non-positve $z$. – Alexey Aug 08 '19 at 21:17
  • @Alexey: It is very much defined (it's essentially what the complex numbers are built on: $(-1)^{1/2}=i$, roughly speaking). – Clayton Aug 08 '19 at 21:18
  • @Clayton, i do not believe there could be a meanigful definition. No, there is no $(-1)^{1/2}$. And there are two distinct square roots of $-1$. – Alexey Aug 08 '19 at 21:19
  • @Alexey: You don't have to believe it for it to be true. – Clayton Aug 08 '19 at 21:21
  • @Clayton, yes, but your claim seems unfounded and probably false. (Unless you are talking about multivalued functions, which is my point.) – Alexey Aug 08 '19 at 21:22
  • @Alexey: I'll leave it at this: roots are not well-defined over complex numbers (hence the need for branch cuts). This does not imply they are not defined, only that we have to state what we mean when taking complex roots. – Clayton Aug 08 '19 at 21:26
  • @Clayton, "we have to state what we mean when taking complex roots" is what i was referring to as not a meaningful definition. (However, if we take all square roots and thus have a multivalued function, this would be meaningful.) – Alexey Aug 08 '19 at 21:29
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Square root is only a single valued function if you restrict yourself to positive real numbers. This is something that people are at least casually aware of when they solve quadratic equations: if you want to solve $x^2=4$, you "take the square root of both sides", but you get two roots, $x=\pm 2$.

When you're not working in a situation where there is a single valued square root function, you need to take the multi-valued aspect more seriously. This means that we no longer have that $\sqrt{z^2}=z$, but rather $\sqrt{z^2}=\{z,-z\}$, and therefore $z\in \sqrt{z^2}$. Another way to patch this is that, if $z$ is real, $|z|=\sqrt{z^2}$. In either case, the identity $\sqrt{z^2}=z$ is only true when $z$ is a positive real number, and cannot be used more generally.

Others have said that the problem is the equality $\sqrt{e^{2\pi i}}=e^{\pi i}$, and while it is true that you only have $\sqrt{e^{2\pi i}}\in \{\pm e^{\pi i}\}$ or $\sqrt{e^{2\pi i}}=|e^{i\pi}|$, it is also true that the problem started with the statement that $1=\sqrt{(-1)(-1)}$, as that is only true is you if you are talking about the single-valued positive square root function, and while you never took the square root of a non-positive number, the manipulations you do make it clear that you want to consider $\sqrt{z}$ more generally.

Aaron
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The first step is wrong already.

In $\mathbb C$ we have $\sqrt1=\{1,-1\}$.

Actually, in $\mathbb R$ we have $\sqrt1=1.$

Michael Rozenberg
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