What is wrong with my 'proof' of $i=1$? 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

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 $\sqrt[4]{x^4}$.
