My note inspired me to make this problem.

Since I said that $p ^ { 2 } = 1$ in that link(my note), but the root of $x ^ { 2 } = 1$ is $x = \pm 1$ in $\mathbb { C }$ .

But, $p ^ { 2 } = 1, ( -p ) ^ { 2 } = 1$ .

So, I enlarged the fundamental theorem of algebra .

And, by that link, we get $n \in \mathbb { N }, p ^ { p } = p ^ { 2n } = 1$ .

$p = 2n$

$2n ^ { 2n } = p ^ { p } = 1$

$( 2n ) ^ { 2n } = 1$

$2n = 1$

$n = \frac { 1 } { 2 }$ .

We found a contradiction.

$p = \frac { 1 } { 0 }$ .

If $n = \frac { 1 } { 2 }$ , then $p = 2n = 1$ .

Then, $1 = \frac { 1 } { 0 }$ .

Therefore, we cannot know whether $p$ is positive or negative like an order of complex numbers.