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So I recently learnt about $i$ and I can't wrap my head around the concept of $i^2=-1$ or that $\sqrt{-1}$ can even exist. Today I was thinking about $i$ again and thought of a "proof" that $i=1$. $$i^4=1 \text{ and } 1^4=1 \text{ so } i^4=1^4$$ $$i^4=1^4 \to \sqrt[4]{i^4}=\sqrt[4]{1^4} \to i=1$$ But if $i=1$, then $i^2 \neq -1$. So I think I must have messed up something in the proof. Can someone point out where this went wrong? I know you can $\text{"prove" }1=2$ by accidentally dividing by $0$ and I suspect something similar is happening.

For anyone else having trouble with complex numbers @mrsamy commented this link and I found it quite helpful: https://www.math.toronto.edu/mathnet/answers/imaginary.html

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    $\begingroup$ In your first sentence, did you intend to write, "I can't wrap my head around the concept of $i^2=−1$ " ? $\endgroup$ – Adam Rubinson Jan 19 at 21:46
  • $\begingroup$ This argument was shown million times. You cannot blindly apply the rules of real numbers to the complex. The flaw is $i^4=1\implies i=\sqrt[4]1$. Anyway, $1=\sqrt[4]1$ is still true. $\endgroup$ – Yves Daoust Jan 19 at 21:46
  • $\begingroup$ "can even exist." Well, nothing in mathematics actually "exists". $\endgroup$ – fleablood Jan 19 at 21:48
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    $\begingroup$ math.toronto.edu/mathnet/answers/imaginary.html $\endgroup$ – user9464 Jan 19 at 21:52
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    $\begingroup$ @mrsamy Thanks a lot for the link. I just saw it and helps a lot $\endgroup$ – Silas Dyck Jan 20 at 19:23
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This has nothing to do with complex numbers. By the same argument, $-1=1$, since$$(-1)^2=1^2\implies\sqrt{(-1)^2}=\sqrt{1^2}\implies-1=1.$$The error lies in assuming that $\sqrt{x^2}=x$. Actually, $\sqrt{x^2}=|x|$. In the case of complex numbers, it's even worst, since every complex number (other than $0$) has four fourth roots. So, the expression $\sqrt[4]z$ doesn't make sense unless and until you decide which fourth root of $z$ you have in mind. Even then, it will often be false that $\sqrt[4]{z^4}=z$.

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    $\begingroup$ Thanks, I figured I must be missing something since I'd tend to assume that people who are more advanced at math than me and say $i$ is a number and the $i=\sqrt{-1}$ are probably correct. $\endgroup$ – Silas Dyck Jan 20 at 16:50
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    $\begingroup$ I'm glad I could help. $\endgroup$ – José Carlos Santos Jan 20 at 17:18
  • $\begingroup$ first equation should have square on both sides, for better mirroring of OP question. Unfortunately that's a 1 char change so I can't make it. $\endgroup$ – Ross Presser Jan 20 at 19:55
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    $\begingroup$ @RossPresser Done. Thank you. $\endgroup$ – José Carlos Santos Jan 20 at 20:00
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Square roots don't work like that. With the same argument you used, you could have done $$ (-1)^2=1^2\ \ \ \implies\ \ \ -1=1. $$ It has nothing to do with complex numbers.

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If this may reconcile you with complex numbers, consider this: a complex number is a pair of reals, and we define

  • addition as $(a,b)+(c,d):=(a+b,c+d)$,

  • multiplication as $(a,b)(c,d):=(ac-bd,ad+bc)$.

Then we accept the "shorthand" conventions $1=(1,0)$ and $i=(0,1)$. It is an easy matter to show that these rules form a consistent arithmetic with the four basic operation, where you can freely use these equivalences.

We also have the consequences

$$i^2=(0,1)(0,1)=(-1,0)=-1$$

and

$$i^4=(-1,0)(-1,0)=(1,0)=1.$$

But that means in no way that $i=1$. It just means that $i$ is one of the solutions of the equation

$$z^4=(1,0)=1.$$

All solutions are (see below)

$$(1,0)=1,\\(0,1)=i,\\(-1,0)=-1,\\(0,-1)=-i.$$

Unless you specify a convention, in the complex we don't know which solution $$\sqrt[4]1$$ denotes.

Final remark: in this discussion, we only used real numbers and the special symbol $i$.

$$z^4=(x,y)^4=(x^4-6x^2y^2+y^4,4yx^3-4xy^3)=(1,0)$$ requires $$x=0\lor y=0\lor x^2=y^2$$

(by cancelling the imaginary part).

By plugging these in the real part, we get $$y^4=1\lor x^4=1\lor -4x^4=-4y^4=1.$$

As $x,y$ are real, the only options are

$$(\pm1,0),(0,\pm1).$$

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Think of the complex numbers as a pair of real numbers $\mathbb{R}\times\mathbb{R}$, but with a really useful property: the product $(x, y)\cdot (0,1) = (-y, x)$. Which shows the rotation characteristic of the number i.

Thinking that way is clear that the number $(0,1)$ has the property of $(0,1)^2 = (-1,0)$.

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The thing is that if $m= n$ then we can conclude $f(m) = f(n)$.

But if $f(m) = f(n)$ we can not conclude $m = n$ unless $f$ is one to one.

If $f$ is not one to one is perfectly possible to have $m \ne n$ but $f(m) = f(n)$.

A simple example is $f(x) = x^2 - 3x +6 $ If $x=2$ we get $f(2) = 4 - 6 + 6 = 4$. And if $x=1$ we get $f(1)=1-3 + 6 = 4$. So $f(1) = f(2) =4$ but $1\ne 2$.

An even easier example is $f(x) = x^4$ and $f(-1) = f(1) = f(i) = f(-i)$ will give us $1^4 = 1$ and $(-1)^4 = 1$ and $i^4 = (i^2)^2 =(-1)^2 =1$ and $(-i)^4 = ((-i)^2)^2=((-1)^2i^2)^2 = (1*(-1))^2 = 1$.

the problem is when sloppy teachers give to impressionable students this INCORRECT definition:

THIS IS WRONG: $\sqrt[k]{m}$ is equal to the $x$ so that $x^k = m$.

The problem is that there are $k$ different $x$ that give $x^k = m$ and so that isn't actually a definition of a single value.

Ex. $\sqrt{16}$ iss the $x$ so that $x^2= 16$. Well $4^2 = 16$ and $(-4)^2 = 16$. So which is it? is $\sqrt{16} = 4$ of is $\sqrt{16} = -4$.

Well, the answer is we define that $\sqrt{m}$ is the positive value $x$ so that $x^2 = m$.

But two things come about from this.

One $\sqrt{x^2} \ne x$. That is just wrong. $\sqrt{x^2} =|x|$ because we don't know that $x$ is positive.

And 2) $i^2 = 1$ is a property of $i$ but that does not mean $\sqrt{-1} = i$. Because $i$ is neither positive nor negative.

Anyway.....

tl;dr

There are four values, all different of $x$ so that $x^4 = 1$. They are $1^4 =1; (-1)^4 = 1; (-i)^4=1$

$\sqrt[4]{x^4} \ne x$. IF $x^4$ is a positive real number then $\sqrt[4]{x^4} = |x|$ and indeed $|1| = |-1| = |i| = |-i|$ but $\sqrt[4]{x^4} \ne x$.

And if $x^4$ is not a positive real number we don't actually have a definition for $\sqrt[4]{x^4}$.

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    $\begingroup$ Sorry, I don't like this explanation much, because you take for granted that $x^4=1$ has four solutions, when the OP is still at a stage that he knows about the two real solutions $\pm1$. Strictly speaking, $i$ is just surrealistic, unless you introduce the imaginary and complex numbers with some formalism. $\endgroup$ – Yves Daoust Jan 19 at 22:20
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    $\begingroup$ Point taken. I was trying explain the error in $\sqrt[4]{x^4} = x$ so $i^4 = 1^4 \implies 1=i$. I was pointing out that $x^4=1$ has more than $1$ solution and not claiming that I was going to prove that it has exactly four in any way. But I did demonstrate that there are at least four. Point taken that $i^2 = -1$ is something tossed in from no-where but I didn't think that was really the issue of the question. $\endgroup$ – fleablood Jan 19 at 23:21
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    $\begingroup$ My comment was influenced by the first sentence of the OP, but no worries. $\endgroup$ – Yves Daoust Jan 20 at 9:03

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